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Applied Calculus IIApplied Calculus II

ProbabilityProbability

Slides subject to changeSlides subject to change

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ProbabilityProbability An An experimentexperiment is a situation involving chance or is a situation involving chance or

probability that leads to results called “outcomes.”probability that leads to results called “outcomes.”

An An outcomeoutcome is the result of a single trial of an is the result of a single trial of an experiment.experiment.

An An eventevent is one or more outcomes of an is one or more outcomes of an experiment.experiment.

ProbabilityProbability is the measure of how likely an event is the measure of how likely an event is.is.

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Probability Equation Probability of an event:

The sample space of an experiment is the set of all possible outcomes of that experiment.

outcomespossibleofnumberTotal

occurcanAeventwaysofNumberAp )(

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Example Experiment: A single 6-sided die is

rolled. What is the probability of each

outcome?   Outcomes:  The possible outcomes of

this experiment are 1, 2, 3, 4, 5 and 6. What is the probability of rolling an

even number? 3/6 or 1/2

This experiment illustrates the difference between an outcome and an event. A single outcome of this experiment is rolling a 1, or rolling a 2, etc. Rolling an even number (2, 4 or 6) − possible outcome of several rolls − is an event, and rolling an odd number (1, 3 or 5) is also an event.

i p(X=i) p(X=i)

1 1/6 0.1667

2 1/6 0.1667

3 1/6 0.1667

4 1/6 0.1667

5 1/6 0.1667

6 1/6 0.1667

Total 6/6 1.0

Sample Sample SpaceSpace

Probability of Probability of Each OutcomeEach Outcome

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Total Number of OutcomesTotal Number of Outcomes Find the probability that when a couple has 3 Find the probability that when a couple has 3

children, they will have exactly 2 boys. Assume children, they will have exactly 2 boys. Assume boys and girls are equally likely.boys and girls are equally likely.

Three correspond to exactly Three correspond to exactly two boys.two boys.

pp((2 boys in 3 births2 boys in 3 births) = ) = 3/8 3/8

= 0.375= 0.375

Sample SpaceSample Space

boy-boy-boyboy-boy-boy

boy-boyboy-boy-girl-girl

boyboy-girl--girl-boyboy

boy-girl-girlboy-girl-girl

girl-girl-boy-boyboy-boy

girl-boy-girlgirl-boy-girl

girl-girl-boygirl-girl-boy

girl-girl-girlgirl-girl-girl

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ProblemProblem A couple is planning to have four children.A couple is planning to have four children. Construct a table of possible gender outcomes.Construct a table of possible gender outcomes. Find the probability of getting exactly two boys Find the probability of getting exactly two boys

and two girls.and two girls. Find the probability that all four children are Find the probability that all four children are

boys.boys.

0.380.38

0.060.06

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Probability RangeProbability Range The probability of an impossible event is 0.The probability of an impossible event is 0.

The probability of an event that is certain to The probability of an event that is certain to occur is 1.occur is 1.

For any event For any event AA, the probability of , the probability of AA is between is between 0 and 1 inclusive, 0 0 and 1 inclusive, 0 ≤ ≤ pp((AA)) ≤ 1.≤ 1.

Impossible Impossible 00

CertainCertain11

Very Very UnlikelyUnlikely

Somewhat Somewhat UnlikelyUnlikely

50-50 50-50 ProbabilityProbability

Rather Rather ProbableProbable

Very Very LikelyLikely

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ProbabilityProbability The complement of event , denoted by The complement of event , denoted by

consists of all outcomes in which event does consists of all outcomes in which event does notnot occur. occur.

Throwing a die, 1 is Throwing a die, 1 is pp(1) = 1/6, (1) = 1/6, pp( ) is 5/6 ( ) is 5/6 (throw a 2, 3, 4, 5, or 6).(throw a 2, 3, 4, 5, or 6).

AAA

1

1)()( ApAp

)(1)( ApAp

)(1)( ApAp

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OddsOdds Expression of likelihood are called odds (such Expression of likelihood are called odds (such

as “50 to 1”)as “50 to 1”) Actual odds in favor, actual odds against ...Actual odds in favor, actual odds against ... Payoff odds established by the racetrack or Payoff odds established by the racetrack or

casino.casino. Odds are 3 to 1 favoring Fred to win the race. In Odds are 3 to 1 favoring Fred to win the race. In

probability terms probability terms pp(Fred’s winning) = ¾ = 0.75.(Fred’s winning) = ¾ = 0.75. If the odds are 3 to 1 against Fred’s winning, If the odds are 3 to 1 against Fred’s winning,

pp(Fred’s winning) = ¼ = 0.25.(Fred’s winning) = ¼ = 0.25.

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Addition RuleAddition Rule A A compoundcompound event is any event combining two or event is any event combining two or

more simple events.more simple events. When two events, A and B, are mutually exclusive, the

probability that A or B will occur is the sum of the probability of each event. 

p(A or B) = p(A) + p(B)

Nicknamed the Nicknamed the “Or – Or Rule.”“Or – Or Rule.” Events are Events are mutually exclusivemutually exclusive if if pp((AA and and BB) = 0, e.g., a ) = 0, e.g., a

die is either 1, or 2, or 3, etc.. A coin is either heads or die is either 1, or 2, or 3, etc.. A coin is either heads or tails.tails.

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Example: Mutually Exclusive A single 6-sided fair die is rolled. What is the probability of rolling a 2 or a 5? Probabilities:  

p(2) = 1/6 p(5) = 1/6

p(2 or 5) = p (2) + p (5) p(2 or 5) =1/6 + 1/6 = 2/6=1/3

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Non-Mutually Exclusive A single card is chosen at random from a

standard deck of 52 playing cards. What is the probability of choosing a king or a

club? A chosen card may be both king and club. This

requires a different equation. The addition causes the king of clubs to be

counted twice, so its probability must be subtracted.

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Non-Mutually Exclusive Probabilities:   p(king or club) = p(king) + p(club) − p(king of clubs)

p(king or club) = 4/52 + 13/52 − 1/52 = 16/52 = 0.308

When two events are non-mutually exclusive, a different addition rule must be used.

p(A or B) = p (A) + p (B) − p (A and B)

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Problems In a pet store, there are 6 puppies, 9 kittens, 4

gerbils and 7 parakeets. If a pet is chosen at random, what is the probability of choosing a puppy or a parakeet?

The probability of a New York teenager owning a skateboard is 0.37, of owning a bicycle is 0.81 and of owning both is 0.36. If a New York teenager is chosen at random, what is the probability that the teenager owns a skateboard or a bicycle?

13/2613/26

0.820.82

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Multiplication RuleMultiplication Rule Two events, Two events, AA and and BB, are , are independentindependent if the  if the

fact that fact that AA occurs does not affect the probability occurs does not affect the probability of of BB occurring. occurring.

When two events, When two events, AA and and BB, are independent, the , are independent, the probability of both occurring is: probability of both occurring is: 

p((AA and and BB) = ) = p ((AA)) • p ((BB))

Nicknamed the Nicknamed the “And-And Rule.”“And-And Rule.”

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Example Experiment:  A coin is tossed and a single 6-sided

die is rolled. Find the probability of a coin landing head−side up

and rolling a 3 on the die.

Probabilities:   p(head) = 1/2 p(roll a 3) = 1/6 p(head and 3) = p (head) • p (roll a 3) = 1/2 · 1/6 = 1/12

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Problem Experiment:  A jar contains 3 red, 5 green, 2

blue and 6 yellow marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen.

What is the probability of choosing a green and a yellow marble?

1/301/30

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Conditional Probability The conditional probability of an event B in

relationship to an event A is the probability that event B occurs given that event A has already occurred.

The notation for conditional probability is

p(B|A)

[pronounced as ”The probability of event B given A”].

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Conditional Probability p(B|A) means the probability of event B given that

event A has already occurred. To find the probability of the two dependent

events, we use a modified version of the multiplication rule.

Multiplication Rule 2:  When two events, A and B, are dependent, the probability of both occurring is: 

p(A and B)  =  p(A)  • p(B|A)

Note

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Example A card is chosen at random from a standard

deck of 52 playing cards. Without replacing it, a second card is chosen.

What is the probability that the first card chosen is a queen and the second card chosen is a jack?

p(A) = 4/52 p(B|A) = 4/51 note this is dependent p(A and B) = 4/52  • 4/51 = 16/2652

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Example On a math test, 5 out of 20 students got an A. If

three students are chosen at random without replacement, what is the probability that all three got an A on the test?

Probabilities:   p(3 and 2 and 1) = 5/20 · 4/19 · 3|18 = 60/6840 = 1/114

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Problem Mr. Parietti needs two students to help him with

a science demonstration for his class of 18 girls and 12 boys. He randomly chooses one student who comes forward. He then chooses a second student from those still seated. What is the probability that both students chosen are girls?

Probabilities:   p(Girl 1 and Girl 2) = p(Girl 1) and p(Girl 2|Girl1) finish

51/14551/145

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Fundamental Counting Principle

How do we count outcomes without listing them? Restaurant offers 4 choices for main dish

(chicken, beef, ham, fish), 3 choices for side dishes (soup, salad, rice) How many days will we be able to order different meals before we have to start repeating our choices?

Principle 1: If k items from one list can be combined with m items from another list, there are km possibilities.

Possible outcomes = 4 x 3 = 12

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Problems  A college offers 7 courses in the morning and 5

in the evening. Find the possible number of choices for the student if he wants to study one course in the morning and one in the evening.

How many different license plates can be made by a state, if each plate is to display three letters followed by three numbers?

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17,576,00017,576,000

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Count Outcomes Principle 2: If you make n choices with

replacement (i.e., an item or value can be selected more than once) from a group of m possibilities, the total number of possibilities is mn.

Seven-letter password: 7 choices each with 26 possibilities. = 267 = 8.03x109.

Chance of someone guessing it is 1/ 8.03x109 = 1.25x10−10.

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Problem A combination lock has 10 digits in

each ring, with 3 rings. How many possible combinations are there?

Another combination lock has 40 numbers, one of which must be selected on each of three turns of the dial. Theoretically, how many possible combinations are there?

Find the number of possible outcomes out of 4 Find the number of possible outcomes out of 4 births, boys and girls being equally probable.births, boys and girls being equally probable. 1,0001,000

64,00064,000

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Ways of Putting Objects in Order If you have n distinct objects, then n! gives the number

of ways of putting the objects in different order, where order does matter. In this case, you have to reduce the number of available choices each time.

The factorial function (symbol: !) just means to multiply a series of descending natural numbers. 0! = 1 1! = 1 2! = 2x1= 2 3! = 3x2x1= 6

)1)...(1(! nnn

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Example What order could 16 pool balls be in? After

choosing, say, number "14" you can't choose it again. So, your first choice would have 16 possibilities, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

If you have five cards, how many ways can you line them up?

5! = 5 x 4 x 3 x 2 x 1 = 120

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Problem A student has a penny, nickel, dime, and

quarter. How many different ways can she line them up?

How many different ways can these 7 Mini-Coopers be lined up?

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5,0405,040

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Permutations

If you select j objects without replacement (such as cards from a deck of cards) from a group of n objects, and you wish to count each ordering separately, then the number of ways of making the selections is given by

This formula gives the number of permutations of n objects taken j at a time, or nPj .

)!(

!

jn

nPjn

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Example If a softball league has 10 teams, how many

different end of the season rankings are possible?  (Assume no ties).

The math is n = 10, j = 10,

10P10 = 10!/(10 − 10)! = 3,628,800

In what order could the first 3 of 16 pool balls be in? The math is n = 16, j = 3,

16P3 = n!/(n − j)! = 16!/(16 − 3)! = 3,360

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Example How many ways can two horses end up in the lead in

a race with eight horses? 8 possibilities for the first place horse. 7 possibilities for the second place horse. Number of possibilities = 8x7 =56

Now, use the equation, with n = 8, j = 2,

nPj = n!/(n − j)! = 8P2 = 8!/6! = 8 x 7 = 56

The probability that you will select the top two in order at random is 1/56 = 0.018 (small).

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Ice Cream Cones

You want variety in your ice cream cones. Store offers 31 flavors. You want a double dip, with two different flavors. How

many options do you have?

Note that vanilla on top, chocolate on bottom is counted as a different option than chocolate on top, vanilla on bottom.

9303031)!29(

!31

)!231(

!31231

xP

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Combinations If you don’t want every possible pair of ice

cream flavors counted twice, there are only 465 combinations.

If order doesn’t matter, it’s a combination. If order does matter, it’s a permutation.

If you select j objects without replacement from a group of n objects, and you do not wish to count each ordering separately, then the number of ways of making the selections is given by

)!(!

!

jnj

nC jn

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Example You want a double dip, with two different flavors,

but you don’t care about order, just that they have two flavors. How many options do you have?

Combination of 31 objects taken 2 at a time.

31C2 (say “31 choose 2”) =

4652

930

!28!2

!31

)!231(!2

!31231

C

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Example How many different committees of 4 students can be

chosen from a group of 15?

15C4 = (15!)/(4! 11!) =1,365

In a conference of 9 schools, how many intraconference football games are played during the season if the teams all play each other exactly once?

There are 9C2 ways of choosing two teams.

9C2 = (9!)/(2! 7!) = 36

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Example You are forming a 12-member swim team from 10

girls and 15 boys. The team must consist of five girls and seven boys. How many 12-member teams are possible?

There are 10C5 of choosing five girls. There are 15C7 ways of choosing seven boys.

Using the Fundamental Counting Principle, there are

10C5 • 15C7 ways choosing five girls and seven boys.

10C5 • 15C7 = (252)(6435) = 1,621,620

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Problem From a group of 40 people, a jury of 12 people is

to be selected. In how many different ways can the jury be selected?

5,586,853,4805,586,853,480

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Problem A university hospital hires 8 people for 4 openings

in the R&D department. Three of the 8 people are women. If all 8 are qualified, in how many ways can the employer fill the four positions if

a) the selection is random?

b) exactly two selections are women? Hint: look at example with the teams. For the women, 3

choose 2, for the men, 5 choose 2.

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