Chapter 6

Probability

VocabularyProbability – the proportion of times the outcome would

occur in a very long series of repetitions (likelihood of an event occuring)

Sample space S – set of all possible outcomesEvent – an outcome or a set of outcomesRelative frequency – likelihood of occurrence (%)Trials – repetitions

Rolling a DieSample Space: S = {1, 2, 3, 4, 5, 6}

Possible Event: A = rolling a 3 P(A) = 1/6

Possible Event: B = rolling an odd P(B) = 3/6

Possible Event: C = number greater than 4 P(C) = 2/6

If we rolled the die 600 times, what type of distribution would we see? Sketch the graph.

Probability Model (probability distribution):

xi 1 2 3 4 5 6

pi 1/6 1/6 1/6 1/6 1/6 1/6

Flipping 2 CoinsSample Space: S = {HH, HT, TH, TT}

Possible Event: 2 heads P(2 heads) = 1/4

Possible Event: one head P(1 head) = 2/4

Probability Model (probability distribution):

If we flip the coin again, does the result of the first flip influence the result of the 2nd flip?

No, events are independent.

pi ¼ ¼ ¼ ¼

xi HH HT TH TT

More VocabularyMultiplication

Principle:

if you can do one task in a number of ways and a second task in b number of ways, then both tasks can be done in a ∙ b number of ways.

Example: What is the total number of outcomes of flipping 4 coins (either all at once or one coin 4 times)?

With Replacement: Draw a card from a deck of cards. Put it back, shuffle and draw again.

Without Replacement: Draw a card from a deck of cards. Set it aside and draw another card.

Example: Select a random digit by drawing numbered slips of paper from a hat. How many 3 digit numbers can you make

a. with replacement?

b. without replacement?

More VocabularyRolling a fair dice is an

example of independent events. Knowledge of what the first roll was has no influence on the next roll.

If I were to randomly select students in class of 100 students without replacement, the probability of you being selected changes with each name called. 1/100, 1/99, 1/98,…0.

Independent Events: knowing one event has occurred does not change the probability of the other event occurring.

Dependent Events: knowing one event has occurred changes the probability of the other event occurring. Conditional Probability

TREE DIAGRAMSflipping a coin and rolling a die

H

T

1

2

3

4

5

6

1

2

3

4

5

6

H1

H2

H3

H4

H5

H6

T1

T2

T3

T4

T5

T6

12 Events (6 x 2)

P(Heads and 1) =

P(Heads and Even) =

P(Tails) =

P(Even) =

ExamplesIn a test of a new package design, you drop a package of 4 glass ornaments.

1. Describe the sample space.2. Given the probability distribution table of

your results, calculate the probabilities:a. Complete the table.

b. What is the probability of breaking at least 2 glasses?

c. What is the probability of breaking no more than 3 glasses?

xi 0 1 2 3 4

pi 0.1 0.2 0.5 0.1 ?

Summary of Rules• 0 P(event) 1• P(S) = 1, where S is Sample Space• P(eventC) = 1 – P(event)• P(A or B) = P(A) + P(B) (if A and B disjoint)• P(A or B) = P(A) + P(B) – P(A and B) • P(A and B) = P(A)P(B) (if A and B independent)

Conditions for Valid Probabilities0 P(event) 1

The probability of an event must be between 0 and 1.P(tails) = .5 P(roll a 4) = 1/6P(today is Friday) = _?_

P(S) = 1Sum of all probabilities of events in a sample space is

1.P(heads) + P(tails) = 1 P(weekday) + P(weekend) = 1

P(A) + P(AC) = 1AC is the complement of event A.

If A = heads, then AC = tails If B = heart, then BC = not heart

More VocabularyTwo events are DISJOINT or MUTUALLY EXCLUSIVE if they do not contain any of the same events and can not occur simultaneously.Joint Events: the simultaneous occurrence of two

events Joint Probability: the probability of a joint event

occurringIndependent Events: knowing one event has occurred does not change the probability of the other event occurring.Dependent Events: knowing one event has occurred changes the probability of the other event occurring.

Disjoint events cannot be independent!

Union of 2 Events: P(A or B) = P(A B)If two events are disjoint (or mutually exclusive)

P(A B) = P(A or B) = P(A) + P(B)

For ALL events A and BP(A B) = P(A or B) = P(A) + P(B) – P(A and B)

∩

∩

∩Intersection of 2 Events: P(A and B) = P(A∩B)

If independent: P(A∩B) = P(A and B) = P(A) P(B)If dependent: Use common sense probability or

conditional probability.If disjoint: P(A and B) = {} or Ø (this is the empty set)

1. Find the probability that you draw either an ace or a red card.

P(Ace or Red) = P(Ace) + P(Red) – P(Red Ace)

= (4/52) + (26/52) – (2/52)

= 28/52

2. Find the probability of rolling 2 sixes.

P( roll 6 and roll 6) = P(roll 6)P(roll 6)

= (1/6)(1/6)

= 1/36

Examples

3. Find the probability of drawing a 7 and a 6 w/o replacement.P( 7and 6) =

= P(7 )P(6 given 7 was drawn)= (1/52)(1/51)

4. Find the probability of rolling a 7 or a 10.P(7 or 10) = P(7) + P(10) – P(7 and 10)

= (6/36) + (3/36)= (9/36)

Examples

For ANY 2 events:P(B|A) = P(A∩B)

P(A)P(A∩B) = P(A) P(B|A)

Two events are independent if:P(B|A) = P(B) (cannot use multiplication rule)

Summary of RulesConditional Probability

Probability of one event GIVEN another has occurred

P(B|A) = P(A∩B) (“probability of B given A”) P(A)

P(A∩B) = P(A) P(B|A)

Or you can use common sense!

same formula!

Conditional Probability

Remember – you

can only use P(A∩B) =

P(A)P(B) if independent!

Conditional Probability ExamplesCommon Sense!

1. Find the probability of drawing a red ball out of a box containing 3 red, 4 blue, and 1 white GIVEN that a blue ball has been drawn and not replaced.

P(R|B) = 3/7

Conditional ProbabilityTree Diagrams!

2. During a one-and-one shooting foul in a basketball game, what is the probability that an 80% free throw shooter makes both baskets?

make (.8)

miss (.2)

1st shot

make (.8)

miss (.2)

2nd shot

2pts = .8*.8=.64

1pt = .8*.2=.16

0pts = .2

Answer: P(making both shots) = .64

3. The probability of having a certain disease is .05. The probability of testing + if you have the disease is .98; the probability of testing + when you do not have it is .10. What is the probability that you have the disease if you test +?

and |

P DP D

P

Conditional Probability ExamplesThe Formula!

.05 .98

.05 .98 .10 .95

.3403

Are the following events disjoint?Are the following events independent?

1. Drawing a jack and a king.

2. Drawing a red card and a king.

3. Drawing an even card and a face card.

4. Drawing an ace and a black card.

5. Drawing an ace and a queen.

6. Drawing a diamond and a red card.