probability ii
DESCRIPTION
Probability II. Denoted by P(Event). Probability. This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely. The relative frequency at which a chance experiment occurs Flip a fair coin 30 times & get 17 heads. - PowerPoint PPT PresentationTRANSCRIPT
Probabilitybull Denoted by P(Event)
outcomes total
outcomes favorable)( EP
This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely
Experimental Probability
bull The relative frequency at which a chance experiment occursndashFlip a fair coin 30 times amp get 17
heads
30
17
Law of Large Numbers
bull As the number of repetitions of a chance experiment increase the difference between the relative frequency of occurrence for an event and the true probability approaches zero
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
Rule 2 Sample spaceIf S is the sample space P(S) = 1
Rule 3 Complement
For any event E P(E) + P(not E) = 1
Rule 4 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)
Ex 1) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Experimental Probability
bull The relative frequency at which a chance experiment occursndashFlip a fair coin 30 times amp get 17
heads
30
17
Law of Large Numbers
bull As the number of repetitions of a chance experiment increase the difference between the relative frequency of occurrence for an event and the true probability approaches zero
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
Rule 2 Sample spaceIf S is the sample space P(S) = 1
Rule 3 Complement
For any event E P(E) + P(not E) = 1
Rule 4 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)
Ex 1) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Law of Large Numbers
bull As the number of repetitions of a chance experiment increase the difference between the relative frequency of occurrence for an event and the true probability approaches zero
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
Rule 2 Sample spaceIf S is the sample space P(S) = 1
Rule 3 Complement
For any event E P(E) + P(not E) = 1
Rule 4 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)
Ex 1) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
Rule 2 Sample spaceIf S is the sample space P(S) = 1
Rule 3 Complement
For any event E P(E) + P(not E) = 1
Rule 4 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)
Ex 1) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Rule 3 Complement
For any event E P(E) + P(not E) = 1
Rule 4 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)
Ex 1) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Rule 4 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)
Ex 1) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 1) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 2) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Independentbull Two events are independent if knowing that
one will occur (or has occurred) does not change the probability that the other occursndash Toss two coins and record the number of heads
Does the probability change for toss 2
ndash Randomly selected a card from a deck and then take another - What is the probability of getting two hearts
Independent
Not independent
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Rule 5 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 3) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 4)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 5) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 6) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 7 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 8) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
given
andA)|P(B
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 9) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Ex 10) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
11) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-
Example 18Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
- PowerPoint Presentation
- Probability
- Experimental Probability
- Law of Large Numbers
- Basic Rules of Probability
- Slide 6
- Slide 7
- Slide 8
- Slide 9
- Independent
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Rule 7 Conditional Probability
- Slide 21
- Slide 22
- Probabilities from two way tables
- Slide 24
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Slide 29
- Slide 30
-