Chapter 7 Sets & Probability

Chapter 7Chapter 7Sets & ProbabilitySets & Probability

Section 7.5Section 7.5

Conditional Probability;Conditional Probability;

Independent EventsIndependent Events

Lie detectors are not admitted as evidence in courtroom Lie detectors are not admitted as evidence in courtroom trials due to the fact they are not 100% reliable. An trials due to the fact they are not 100% reliable. An experiment was conducted in which a group of suspects experiment was conducted in which a group of suspects was instructed to lie or tell the truth to a set of questions, was instructed to lie or tell the truth to a set of questions, and a group of polygraph experts, along with the polygraph and a group of polygraph experts, along with the polygraph (lie detector), judged whether the suspect was telling the (lie detector), judged whether the suspect was telling the truth or not. The results are tabulated below.truth or not. The results are tabulated below.

Experts’ JudgmentExperts’ Judgment TruthTruth LieLie

TruthTruth 9393 1111

LieLie 77 8989

Totals Totals 100100 100100

Suspects’ Answers

Estimate the probability that:Estimate the probability that:

a.) there will be a miscarriage of justice.a.) there will be a miscarriage of justice.

18 / 200 = .0918 / 200 = .09

b.) a suspect is a liar and gets away with the lie.b.) a suspect is a liar and gets away with the lie.

11 / 200 = .05511 / 200 = .055

c.) the experts’ judgment is correct.c.) the experts’ judgment is correct.

182 / 200 = .91182 / 200 = .91



LieLie 77 8989


Suspects’ Answers

Estimate the probability that:Estimate the probability that:

d.) a suspect is telling the truthd.) a suspect is telling the truth

100 / 200 = .5100 / 200 = .5

e.) a suspect is found to be honeste.) a suspect is found to be honest

104 / 200 = .52104 / 200 = .52

f.) a suspect who is telling the truth is found to be honest by the f.) a suspect who is telling the truth is found to be honest by the experts.experts.

93 / 100 = .9393 / 100 = .93



LieLie 77 8989


Suspects’ Answers

Conditional ProbabilityConditional ProbabilityA probability problem in which the sample space is A probability problem in which the sample space is reduced by known, or given, information is called a reduced by known, or given, information is called a conditional probabilityconditional probability. In other words, a conditional . In other words, a conditional probability exists when the sample space has been probability exists when the sample space has been limited to only those outcomes that fulfill a certain limited to only those outcomes that fulfill a certain condition.condition.

In a newspaper poll concerning violence on television, 600 In a newspaper poll concerning violence on television, 600 people were asked, “What is your opinion of the amount of people were asked, “What is your opinion of the amount of violence on prime-time television – is there too much violence on prime-time television – is there too much violence on television?” Their responses are indicated in violence on television?” Their responses are indicated in the table below.the table below.

YesYes NoNo Don’t KnowDon’t Know TotalTotal

MenMen 162162 9595 2323 280280

WomenWomen 256256 4545 1919 320320

TotalTotal 418418 140140 4242 600600

YesYes NoNo Don’t KnowDon’t Know TotalTotal

MenMen 162162 9595 2323 280280

WomenWomen 256256 4545 1919 320320

TotalTotal 418418 140140 4242 600600

Too Much Violence on Television?

Use the table to find the probabilities below.Use the table to find the probabilities below.P (Y)P (Y)

P (M)P (M)

P (Y | M)P (Y | M)

P (M | Y)P (M | Y)

P (Y P (Y M) M)

P (M P (M Y) Y)

A pair of dice is rolled. Find the probabilities of the given A pair of dice is rolled. Find the probabilities of the given events.events.

a.) The sum is 12a.) The sum is 12

b.) The sum is 12, given that the sum is evenb.) The sum is 12, given that the sum is even

c.) The sum is 12, given that the sum is oddc.) The sum is 12, given that the sum is odd

d.) The sum is even, given that the sum is 12d.) The sum is even, given that the sum is 12

e.) The sum is 4, given that the sum is less than 6e.) The sum is 4, given that the sum is less than 6

f.) The sum is less than 6, given that the sum is 4f.) The sum is less than 6, given that the sum is 4

A single die is rolled. Find the probabilities of the A single die is rolled. Find the probabilities of the given events.given events.

a.) rolling a 5a.) rolling a 5

b.) rolling a 5, given that the number rolled is oddb.) rolling a 5, given that the number rolled is odd

c.) rolling an odd number, given that a 5 was rolledc.) rolling an odd number, given that a 5 was rolled

A pair of dice is rolled. Find the A pair of dice is rolled. Find the probabilities of the given events.probabilities of the given events.

a.) sum is 10a.) sum is 10

b.) sum is 10, given the sum is evenb.) sum is 10, given the sum is even

c.) sum is 7, given the sum is oddc.) sum is 7, given the sum is odd

d.) sum is even, given the sum is 8d.) sum is even, given the sum is 8

Product Rule of ProbabilityProduct Rule of Probability

The Product Rule gives a method for finding the probability that events E and F both occur.

ExampleExample

Two cards are drawn without replacement Two cards are drawn without replacement from a standard deck of 52 cards.from a standard deck of 52 cards.

a.) Find the probability of getting a King a.) Find the probability of getting a King

followed by an Ace.followed by an Ace.

b.) Find the probability of drawing a 7 and b.) Find the probability of drawing a 7 and

a Jack.a Jack.

c.) Find the probability of drawing two Aces.c.) Find the probability of drawing two Aces.

The Nissota Automobile Company buys emergency The Nissota Automobile Company buys emergency flashers from two different manufacturers: one in flashers from two different manufacturers: one in Arkansas and one in Nevada. Arkansas and one in Nevada. •Thirty-nine percent of its turn-signal indicators are Thirty-nine percent of its turn-signal indicators are purchased from the Arkansas manufacturer, and the purchased from the Arkansas manufacturer, and the rest are purchased from the Nevada manufacturer. rest are purchased from the Nevada manufacturer. •Two percent of the Arkansas turn-signal indicators are Two percent of the Arkansas turn-signal indicators are defective, and 1.7% of the Nevada indicators are defective, and 1.7% of the Nevada indicators are defective.defective.

a.) What percent of the defective indicators are made a.) What percent of the defective indicators are made in Arkansas?in Arkansas?

b.) What percent of the defective indicators are made b.) What percent of the defective indicators are made in Nevada?in Nevada?

Manufacturer Indicator

Arkansas

Nevada

Defective

Defective

Not Defective

Not Defective

.39

.61

.02

.98

. 017

.983


Arkansas

Nevada

Defective

Defective

Not Defective

Not Defective

.39

.61

.02

.98

. 017

.983

a.) What percent of the defective indicators are made in Arkansas?a.) What percent of the defective indicators are made in Arkansas?

P(A D) = P(A) ∙ P(D | A)P(A)

P(D | A)

P(A D) = .39 ∙ .02 = .0078

____.0078____ .0078 + .01037 ≈ .4293 ≈ 42.93%=P ( A | D) =

P (A D ) P(D)

P(N D) = P(N) ∙ P(D | N)P(N)

P(D | N)

P(N D) = .61 ∙ .017 = .01037


Arkansas

Nevada

Defective

Defective

Not Defective

Not Defective

.39

.61

.02

.98

. 017

.983

b.) What percent of the defective indicators are made in Nevada?b.) What percent of the defective indicators are made in Nevada?

P ( N | D) = 1 - P ( A | D) = 1 - .4293 = .5707 = 57.07%

(Using Complement Rule)

Two cards are dealt from a full deck of 52. Find the Two cards are dealt from a full deck of 52. Find the probabilities of the given events. (Hint: Make a tree diagram, probabilities of the given events. (Hint: Make a tree diagram, labeling each branch with the appropriate probabilities.)labeling each branch with the appropriate probabilities.)

a.) The first card is a king.a.) The first card is a king.

b.) Both cards are kingsb.) Both cards are kings

c.) The second card is a king, given that the first card was c.) The second card is a king, given that the first card was a a king.king.

d.) The second card is a kingd.) The second card is a king

A coin is flipped twice in succession. Find the probabilities A coin is flipped twice in succession. Find the probabilities of the given events.of the given events.

a.) Both tosses result in tailsa.) Both tosses result in tails

b.) The second toss is tails given the first toss is tailsb.) The second toss is tails given the first toss is tails

c.) The second toss results in tailsc.) The second toss results in tails

d.) Getting tails on one toss and heads on the otherd.) Getting tails on one toss and heads on the other

Independent EventsIndependent Events

Two events are Two events are independentindependent if the probability of if the probability of one event one event does notdoes not depend / affect the depend / affect the probability (or occurrence) of the other event.probability (or occurrence) of the other event.

In other words, knowing F does not affect E’s probability.

Product Rule for Product Rule for Independent EventsIndependent Events

Dependent EventsDependent Events

Two events are Two events are dependentdependent if the if the probability of one event probability of one event doesdoes affect the affect the probability (or likelihood of occurrence) of probability (or likelihood of occurrence) of the other event.the other event.

Two events Two events EE and and FF are dependent if are dependent if P(P(EE | | FF) ) P(E) or P(E) or P(P(FF | | EE) ) P( P(FF) )

Use your own personal experience or probabilities to determine Use your own personal experience or probabilities to determine whether the following events E and F are mutually exclusive and/or whether the following events E and F are mutually exclusive and/or independent.independent.

a.) E is the event “being a doctor” and F is the event “being a.) E is the event “being a doctor” and F is the event “being a woman”.a woman”.

b.) E is the event “it’s raining” and F is the event “it’s b.) E is the event “it’s raining” and F is the event “it’s sunny”.sunny”.

c.) E is the event “being single” and F is the event “being c.) E is the event “being single” and F is the event “being married”.married”.

d.) E is the event “having naturally blond hair” and F is the d.) E is the event “having naturally blond hair” and F is the event “having naturally black hair”.event “having naturally black hair”.

e.) If a die is rolled once, and E is the event “getting a 4” e.) If a die is rolled once, and E is the event “getting a 4” and F is the event “getting an odd number”.and F is the event “getting an odd number”.

f.) If a die is rolled once, and E is the event “getting a 4” f.) If a die is rolled once, and E is the event “getting a 4” and F is the event “getting an even number”.and F is the event “getting an even number”.

A skateboard manufacturer buys 23% of its ball bearing from A skateboard manufacturer buys 23% of its ball bearing from a supplier in Akron, 38% from one in Atlanta, and the rest from a supplier in Akron, 38% from one in Atlanta, and the rest from a supplier in Los Angeles. Of the ball bearings from Akron, a supplier in Los Angeles. Of the ball bearings from Akron, 4% are defective; 6.5% of those from Atlanta are defective; 4% are defective; 6.5% of those from Atlanta are defective; and 8.1% of those from Los Angeles are defective.and 8.1% of those from Los Angeles are defective.

a.) Find the probability that a ball bearing is defective.a.) Find the probability that a ball bearing is defective.

b.) Are the events “defective” and “from the Los Angeles b.) Are the events “defective” and “from the Los Angeles supplier” independent? Show mathematical supplier” independent? Show mathematical justification.justification.

c.) Are the events “defective” and “from the Atlanta c.) Are the events “defective” and “from the Atlanta supplier” supplier” independent? Show mathematical justification.independent? Show mathematical justification.

d.) What conclusion can you draw?d.) What conclusion can you draw?

Documents

Chapter 7 Sets & Probability