Let’s consider this problem…The table below gives the proportion of time that the gerbilspends in each compartment.

Compartment

Proportion

A

0.25

B

0.20

C

0.35

D

0.30

9.(a) Is this a valid probability function? Explain.

No. 0.25 + 0.20 + 0.35 + 0.30 = 1.1,No. 0.25 + 0.20 + 0.35 + 0.30 = 1.1,which does not equal 1.which does not equal 1.

9.(b) Is there a problem with Alrik’s reasoning? Explain

Yes. Since the gerbil must always beYes. Since the gerbil must always bein exactly one of the four rooms, thein exactly one of the four rooms, the

proportions must add up to 1.proportions must add up to 1.

Let’s consider this problem…The table below gives the proportion of time that the gerbilspends in each compartment.

Compartment

Proportion

A

0.400.40

B

0.300.30

C

0.200.20

D

0.100.10

10. Suppose Alrik determines that his gerbil spends time in thefour compartments A, B, C, and D in the ratio 4:3:2:1. Whatproportions should he fill in the table above? Is this a validprobability function?

Since 4 + 3 + 2 + 1 = 10, we can divide eachSince 4 + 3 + 2 + 1 = 10, we can divide eachnumber in the ratio by 10 and get thenumber in the ratio by 10 and get the

proportions relative to the whole.proportions relative to the whole.

Sec. 9.3b

Some Useful Diagrams for ProbabilitySome Useful Diagrams for Probability

Venn DiagramVenn Diagram – good for visualizing – good for visualizingrelationships between events in a givenrelationships between events in a givensample spacesample space

Tree DiagramTree Diagram – good for visualizing – good for visualizingproblems that rely on the Multiplicationproblems that rely on the MultiplicationPrinciple of Probability.Principle of Probability.

In a large high school, 54% of the students are girls and 62%of the students play sports. Half of the girls at the school playsports.(a) What percentage of the students who play sports are boys?

(b) If a student is chosen at random, what is the probability that it is a boy who does not play sports?

Venn Diagram

Girls Sports

0.270.270.270.27 0.350.35

0.110.11

(a) Ratio of boys who play sports to all students who play sports:

0.35

0.62= 0.5645

About 56.45% of theAbout 56.45% of the students who play sportsstudents who play sports are boysare boys

In a large high school, 54% of the students are girls and 62%of the students play sports. Half of the girls at the school playsports.(a) What percentage of the students who play sports are boys?

(b) If a student is chosen at random, what is the probability that it is a boy who does not play sports?

Venn Diagram

Girls Sports

0.270.270.270.27 0.350.35

0.110.11

(b) A boy that also does not play sports is represented outside of both circles:

0.11

There is an 11% chanceThere is an 11% chance that a student chosen atthat a student chosen at random is a boy who doesrandom is a boy who does not play sportsnot play sports

Two identical cookie jars are on a counter. Jar A contains 2chocolate chip and 2 peanut butter cookies, while jar B contains1 chocolate chip cookie. We select a cookie at random. Whatis the probability that it is a chocolate chip cookie?

There are 5 total cookies, 3 of which are chocolate chip…

P(chocolate chip) = 3/5, right???

No the fact that the cookies are in different jars means thatthey are not equally likely outcomes… (i.e., the chocolate chipcookie in jar B is more likely to be chosen than any singlecookie in jar A)

We need to visualize this experiment as a two-step process:First choose a jar, then choose a cookie from that jar!!!

Two identical cookie jars are on a counter. Jar A contains 2chocolate chip and 2 peanut butter cookies, while jar B contains1 chocolate chip cookie. We select a cookie at random. Whatis the probability that it is a chocolate chip cookie?

JarA

JarB

CC

PB

PB

CC

CC

Tree Diagram

0.5

0.5

1

0.25

0.25

0.25

0.25

0.1250.125

0.1250.125

0.1250.125

0.1250.125

0.50.5

The event “chocolate chip”is a set that contains threeoutcomes. Add theirprobabilities:

P(chocolate chip) =P(chocolate chip) =

0.125 + 0.125 + 0.50.125 + 0.125 + 0.5

= 0.75= 0.75

Conditional ProbabilityConditional ProbabilityOur previous example is one of conditional probability, sincethe “cookie” outcome is dependent on the “jar” outcome.

Notation:

P(A|B)P(A|B) • Read “P of A given B”

• Meaning “the probability of the event A, given that event B occurs”

Previous Example:

P(chocolate chip | jar A) =2

4P(chocolate chip | jar B) = 1

Multiplication Principle of Probability:

P(A and B) = P(A) x P(B|A)P(A and B) = P(A) x P(B|A)

Conditional ProbabilityConditional Probability

Conditional Probability Formula:Conditional Probability Formula:

If the event B depends on the event A, then

P (B | A) =P(A and B)

P(A)

Conditional ProbabilityConditional ProbabilitySuppose we have drawn a cookie at random from oneof the jars described in the previous example. Giventhat it is chocolate chip, what is the probability that itcame from jar A?

P(jar A | chocolate chip) =

P(jar A and chocolate chip)

P(chocolate chip)=

(1/2)(2/4)

0.75=

1

3

Guided PracticeGuided PracticeA and B are events in a sample space S such that P(A) = 0.7,P(B) = 0.4, and P(A and B) = 0.2.

A B

0.20.20.5

0.1

1. Find the probability that A occurs but B does not. P = 0.5P = 0.5

2. Find the probability that B occurs but A does not. P = 0.2P = 0.2

3. Find the probability that neither A nor B occurs. P = 0.1P = 0.1

Guided PracticeGuided PracticeA and B are events in a sample space S such that P(A) = 0.7,P(B) = 0.4, and P(A and B) = 0.2.

A B

0.20.20.5

0.1

4. Are events A and B independent? (That is, does P(A | B) = P(A))? NONO

P(A | B) =P(A | B) =P(A and B)P(A and B)

P(B)P(B)

0.20.2

0.40.4== = 0.5 = P(A)= 0.5 = P(A)

Guided PracticeGuided PracticeIf the school cafeteria serves meat loaf, there is a 70% chancethat they will serve peas. If they do not serve meat loaf, there isa 30% chance that they will serve peas anyway. The studentsknow that meat loaf will be served exactly once during the 5-dayweek, but they do not know which day. If tomorrow is Monday,what is the probability that

(a) the cafeteria serves meat loaf?

(b) the cafeteria serves meat loaf and peas?

(c) the cafeteria serves peas?

ML

NML

0.2

0.8

0.7

0.3

P

NP

P

NP

0.3

0.7

P = 0.2P = 0.2P = 0.14P = 0.14

P = 0.38P = 0.38

0.14

0.06

0.24

0.56

Binomial Distributions

Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(a) all 3’s.

There is a probability 1/6 of rolling a three each time…

Multiplication Principle:

P(rolling 3 four times) = (1/6) 4 0.000772

Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(b) no 3’s.

There is a probability 5/6 of rolling a non-three each time…

Multiplication Principle:

P(rolling non-3 four times) = (5/6)4 0.482

Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(c) exactly two 3’s.

Probability of rolling two 3’s followed by two non-3’s:

(1/6) (5/6)2 0.019292

However, there are other outcomes to consider!!!

The two 3’s could occur anywhere among the four rolls In how many ways? 4

62

ways

Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(c) exactly two 3’s.

So we have 6 possible outcomes, each with probability:

(1/6) (5/6)2 2

P(exactly two 3’s) = 2 241 6 5 6 0.11574

2

Let’s start with a familiar experiment

The forms of these three answers should look familiar…Let’s let p = 1/6 and q = 5/6:

P(four 3’s) = p4

P(no 3’s) = q 4

P(two 3’s) = p q24

22

These are three terms in theexpansion of (p + q) !!!4

Let’s start with a familiar experiment

The forms of these three answers should look familiar…Let’s let p = 1/6 and q = 5/6:

The terms in the expansion

4 4 3 1 2 2 1 3 44 6 4p q p p q p q p q q give the exact probabilities of 4, 3, 2, 1, and 0 threes (respectively)when tossing a fair die four times!!!

For this reason, such an example yields what is called a

Binomial Probability DistributionBinomial Probability Distribution

Theorem: Binomial DistributionTheorem: Binomial DistributionSuppose an experiment consists of n independent repetitions ofan experiment with two outcomes, called “success” and “failure.”Let P(success) = p and P(failure) = q. (Note that q = 1 – p.)

Then the terms in the binomial expansion of (p + q) give therespective probabilities of exactly n, n – 1,…, 2, 1, 0 successes.The distribution is shown on the following slide…

n

Number of successes out ofn independent repetitions Probability

n np

1n 1

1nnp q

n

r r n rnp q

r

1 1

1nn

pq

0 nq

More Practice Problems

Suppose Michael makes 90% of his free throws. If he shoots20 free throws, and if his chance of making each one isindependent of the other shots, what is the probability that hemakes (a) all 20, (b) exactly 18, and (c) at least 18?

This situation yields a binomial probability distribution,with p = success = 0.9, and q = failure = 0.1.

(a) P(20 successes) = 200.9 0.12158

(b) P(18 successes) = 18 2200.9 0.1 0.28518

18

More Practice Problems

Suppose Michael makes 90% of his free throws. If he shoots20 free throws, and if his chance of making each one isindependent of the other shots, what is the probability that hemakes (a) all 20, (b) exactly 18, and (c) at least 18?

(c) P(at least 18 successes) =

18 2 19 2020 200.9 0.1 0.9 0.1 0.9

18 19

P(18) + P(19) + P(20)

0.6769

More Practice Problems

A new medical test for a deadly virus is 0.7% likely to produce afalse positive result (i.e., the test indicates the presence of thevirus when it is not actually present). If 60 individuals are tested,what is the probability that (a) there will be 3 false positives,and (b) at least one false positive?

(a) P(3 false) = 3 57600.007 0.993

3

0.00786

Here, p = 0.007 and q = 0.993

More Practice Problems

A new medical test for a deadly virus is 0.7% likely to produce afalse positive result (i.e., the test indicates the presence of thevirus when it is not actually present). If 60 individuals are tested,what is the probability that (a) there will be 3 false positives,and (b) at least one false positive?

(b) P(at least 1 false) = 1 – P(no false)

601 0.993 0.344

Here, p = 0.007 and q = 0.993

More Practice Problems

Ten dimes, dated 1990 through 1999 are tossed. Find theprobability of each of the following events.

P(H-T-T-T-T-T-T-T-T-T)

101

2

1. Heads on the 1990 dime only.

10.0976%

1024

P(T-H-T-T-T-T-H-T-T-T)

101

2

2. Heads on the 1991 and 1996 dimes only.

10.0976%

1024

More Practice Problems

Ten dimes, dated 1990 through 1999 are tossed. Find theprobability of each of the following events.

P(9H and 1T)

10

10 1

1

2C

3. Heads on all but one dime.

100.976%

1024

4. Exactly two heads.

P(2H and 8T)

10

10 2

1

2C

45

4.39%1024