math 1300: section 8-3 conditional probability, intersection, and independence

Conditional ProbabilityIntersection of Events: Product Rule

Probability TreesIndependent Events

Summary

Math 1300 Finite MathematicsSection 8-3: Conditional Probability, Intersection, and

Independence

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

Summary

Conditional Probability

Conditional probability is the probability of an eventoccuring given that another event has already occured.

The conditional probability of event A occuring, given thatevent B has occured is denoted P(A|B), and is read as“probability of A, given B.”

Summary

Conditional probability is the probability of an eventoccuring given that another event has already occured.The conditional probability of event A occuring, given thatevent B has occured is denoted P(A|B), and is read as“probability of A, given B.”

Summary

DefinitionFor events A and B in a arbitrary sample space S, we definethe conditional probability of A given B by

P(A|B) =P(A ∩ B)

P(B)P(B) 6= 0

Summary

0.15.35 .25

P(A ∩ B) = 0.15“out of all outcomes in thesample space, 15% ofthem are in A ∩ B.

P(A|B) asks: out of alloutcomes in B, whatproportion of them are alsoin A.So, we can think of theconditional probability asrestricting the samplespace.

P(A|B) =P(A ∩ B)

=0.150.4

= 0.375

Summary

0.15.35 .25

P(A ∩ B) = 0.15

“out of all outcomes in thesample space, 15% ofthem are in A ∩ B.

P(A|B) =P(A ∩ B)

=0.150.4

= 0.375

Summary

0.15.35 .25

P(A|B) =P(A ∩ B)

=0.150.4

= 0.375

Summary

0.15.35 .25

P(A|B) asks: out of alloutcomes in B, whatproportion of them are alsoin A.

So, we can think of theconditional probability asrestricting the samplespace.

P(A|B) =P(A ∩ B)

=0.150.4

= 0.375

Summary

0.15.35 .25

P(A|B) =P(A ∩ B)

=0.150.4

= 0.375

Summary

0.15.35 .25

P(A|B) =P(A ∩ B)

=0.150.4

= 0.375

Summary

Example: Suppose events A, B, D and E have probabilities asgiven in the table.

A B TotalsD 0.2 0.4 0.6E 0.28 0.12 0.4

Totals 0.48 0.52 1

P(B|D) =P(B ∩ D)

P(A|E) =P(A ∩ E)

0.280.4

Summary

A B TotalsD 0.2 0.4 0.6E 0.28 0.12 0.4

Totals 0.48 0.52 1

P(B|D) =P(B ∩ D)

P(A|E) =P(A ∩ E)

0.280.4

Summary

A B TotalsD 0.2 0.4 0.6E 0.28 0.12 0.4

Totals 0.48 0.52 1

P(B|D) =P(B ∩ D)

P(A|E) =P(A ∩ E)

0.280.4

Summary

A B TotalsD 0.2 0.4 0.6E 0.28 0.12 0.4

Totals 0.48 0.52 1

P(B|D) =P(B ∩ D)

P(A|E)

=P(A ∩ E)

0.280.4

Summary

A B TotalsD 0.2 0.4 0.6E 0.28 0.12 0.4

Totals 0.48 0.52 1

P(B|D) =P(B ∩ D)

P(A|E) =P(A ∩ E)

0.280.4

Summary

Example: In a survey of 255 people, the following data wereobtained relating gender to political orientation:

Republican (R) Democrat (D) Ind. (I) TotalMale (M) 79 22 17 118

Female (F) 40 77 20 137Total 119 99 37 255

A person is randomly selected. What is the probability that:

The person is male?

P(M) =n(M)

118255

≈ .463

The person is male and a Democrat?

P(M ∩ D) =n(M ∩ D)

≈ .086

Summary

Female (F) 40 77 20 137Total 119 99 37 255

A person is randomly selected. What is the probability that:The person is male?

P(M) =n(M)

118255

≈ .463

P(M ∩ D) =n(M ∩ D)

≈ .086

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) =n(M)

118255

≈ .463

P(M ∩ D) =n(M ∩ D)

≈ .086

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) =n(M)

118255

≈ .463

P(M ∩ D) =n(M ∩ D)

≈ .086

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) =n(M)

118255

≈ .463

P(M ∩ D) =n(M ∩ D)

≈ .086

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) = .46P(M ∩ D) = 0.086

a Democrat?

P(D) =n(D)

≈ .39

Male, given that the person is a Democrat?

P(M|D) =P(M ∩ D)

0.086.39

≈ .22

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) = .46P(M ∩ D) = 0.086a Democrat?

P(D) =n(D)

≈ .39

P(M|D) =P(M ∩ D)

0.086.39

≈ .22

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) = .46P(M ∩ D) = 0.086a Democrat?

P(D) =n(D)

≈ .39

P(M|D) =P(M ∩ D)

0.086.39

≈ .22

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) = .46P(M ∩ D) = 0.086a Democrat?

P(D) =n(D)

≈ .39

P(M|D) =P(M ∩ D)

0.086.39

≈ .22

Summary

Female (F) 40 77 20 137Total 119 99 37 255

P(M) = .46P(M ∩ D) = 0.086a Democrat?

P(D) =n(D)

≈ .39

P(M|D) =P(M ∩ D)

0.086.39

≈ .22

Summary

NoteNote that for sample spaces with equally likely outcomes

P(A|B) =n(A ∩ B)

So, for example, from the previous problem we could have done

P(M|D) =n(M ∩ D)

2299≈ .22

Summary

NoteNote that for sample spaces with equally likely outcomes

P(A|B) =n(A ∩ B)

So, for example, from the previous problem we could have done

P(M|D) =n(M ∩ D)

2299≈ .22

Summary

Intersection of Events: Product Rule

In the previous section we sometimes used the additionprinciple

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

to find P(A ∩ B).

The conditional probability formula gives another methodfor finding P(A ∩ B) when appropriate, called the ProductRule.

Summary

Intersection of Events: Product Rule

In the previous section we sometimes used the additionprinciple

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

to find P(A ∩ B).The conditional probability formula gives another methodfor finding P(A ∩ B) when appropriate, called the ProductRule.

Summary

P(A|B) =P(A ∩ B)

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

(P(A ∩ B)

)P(B)P(A|B) = P(A ∩ B)

P(A ∩ B) = P(A|B)P(B)

Similarly, the formula

P(B|A) = P(B ∩ A)P(A)

gives P(A ∩ B) = P(B|A)P(A).

Summary

P(A|B) =P(A ∩ B)

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

(P(A ∩ B)

P(B)P(A|B) = P(A ∩ B)

P(A ∩ B) = P(A|B)P(B)

P(B|A) = P(B ∩ A)P(A)

Summary

P(A|B) =P(A ∩ B)

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

(P(A ∩ B)

)P(B)P(A|B) = P(A ∩ B)

P(A ∩ B) = P(A|B)P(B)

P(B|A) = P(B ∩ A)P(A)

Summary

P(A|B) =P(A ∩ B)

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

(P(A ∩ B)

)P(B)P(A|B) = P(A ∩ B)

P(A ∩ B) = P(A|B)P(B)

P(B|A) = P(B ∩ A)P(A)

Summary

P(A|B) =P(A ∩ B)

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

(P(A ∩ B)

)P(B)P(A|B) = P(A ∩ B)

P(A ∩ B) = P(A|B)P(B)

P(B|A) = P(B ∩ A)P(A)

Summary

TheoremFor events A and B with nonzero probabilities in a samplespace S,

P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)

and we can use either P(A)P(B|A) or P(B)P(A|B) to computeP(A ∩ B).

Summary

Example: A computer chip manufacturer produces 45% of itschips at Plant A and at and the remainder at Plant B. However,1% of the chips produced at Plant A are defective, while 0.5%of the chips produced at Plant B are defective. What is theprobability that a randomly chosen computer chip produced bythis manufactuer is defective and was produced at Plant A?

Let A be the event that a chip was produced at plant A. Let Drepresent the event that a chip is defective. Then,

P(A) = 0.45 and P(A′) = 0.55

(Note that A′ represents the event that a chip was produced atplant B.)

Summary

Let A be the event that a chip was produced at plant A. Let Drepresent the event that a chip is defective.

P(A) = 0.45 and P(A′) = 0.55

Summary

Let A be the event that a chip was produced at plant A. Let Drepresent the event that a chip is defective. Then,

P(A) = 0.45 and P(A′) = 0.55

Summary

The statement “1% of the chips produced at Plant A aredefective, gives the conditional probability

P(D|A) = 0.01

Similarly, the statement ”0.5% of the chips produced at Plant Bare defective“ gives

P(D|A′) = 0.005

The question

What is the probability that a randomly chosencomputer chip produced by this manufactuer isdefective and was produced at Plant A?

is asking for P(D ∩ A).

Summary

P(D|A) = 0.01

P(D|A′) = 0.005

The question

Summary

P(D|A) = 0.01

P(D|A′) = 0.005

The question

Summary

P(D|A) = 0.01

P(D|A′) = 0.005

The question

Summary

By the product rule

P(D ∩ A) = P(D|A)P(A)

P(D ∩ A) = (0.01)(0.45) = 0.0045

So, the probability that a randomly chosen computer chipproduced by this manufactuer is defective and was produced atPlant A is 0.45%.

Summary

By the product rule

P(D ∩ A) = P(D|A)P(A)P(D ∩ A) = (0.01)(0.45) = 0.0045

Summary

By the product rule

P(D ∩ A) = P(D|A)P(A)P(D ∩ A) = (0.01)(0.45) = 0.0045

Summary

Probability Trees

Probability TreesWe used tree diagrams in the previous chapter to help uscount the number of combined outcomes in a sequence ofexperiments.

Similarly, we can use tree diagrams to help us compute theprobabilities of combined outcomes in a sequence ofexperiments.A seqence of experiments where the outcome of eachexperiment is not certain is called a stochastic process.

Summary

Probability Trees

Probability TreesWe used tree diagrams in the previous chapter to help uscount the number of combined outcomes in a sequence ofexperiments.Similarly, we can use tree diagrams to help us compute theprobabilities of combined outcomes in a sequence ofexperiments.

A seqence of experiments where the outcome of eachexperiment is not certain is called a stochastic process.

Summary

Probability Trees

Probability TreesWe used tree diagrams in the previous chapter to help uscount the number of combined outcomes in a sequence ofexperiments.Similarly, we can use tree diagrams to help us compute theprobabilities of combined outcomes in a sequence ofexperiments.A seqence of experiments where the outcome of eachexperiment is not certain is called a stochastic process.

Summary

Constructing Probability TreesDraw a tree diagram corresponding to all combinedoutcomes of the sequence of experiments. Label eachnode.Assign a probability to each tree branch.Use the results from the first two steps to answer variousquestions related to the sequence of experiements as awhole.

Summary

Example: Suppose that 5% of all adults over 40 have diabetes.A certain physician correctly diagnoses 90% of all adults over40 with diabetes as having the disease and incorrectlydiagnoses 3% of all adults over 40 without diabetes as havingthe disease.

(a) Let

A = the event ”has diabetes", and letB = the event ”is diagnosed with diabetes."

Draw a tree diagram which models this scenario and label itwith appropriate probabilities.

Summary

Example: Suppose that 5% of all adults over 40 have diabetes.A certain physician correctly diagnoses 90% of all adults over40 with diabetes as having the disease and incorrectlydiagnoses 3% of all adults over 40 without diabetes as havingthe disease.

(a) Let

A = the event ”has diabetes", and letB = the event ”is diagnosed with diabetes."

Draw a tree diagram which models this scenario and label itwith appropriate probabilities.

Summary

A = ”has diabetes" D = ”is diagnosed with diabetes."

”Suppose that 5% of alladults over 40 havediabetes.“

”A certain physiciancorrectly diagnoses 90% ofall adults over 40 withdiabetes as having thedisease. . . “and incorrectly diagnoses3% of all adults over 40without diabetes as havingthe disease.

Summary

”Suppose that 5% of alladults over 40 havediabetes.“”A certain physiciancorrectly diagnoses 90% ofall adults over 40 withdiabetes as having thedisease. . . “

and incorrectly diagnoses3% of all adults over 40without diabetes as havingthe disease.

Summary

”Suppose that 5% of alladults over 40 havediabetes.“”A certain physiciancorrectly diagnoses 90% ofall adults over 40 withdiabetes as having thedisease. . . “and incorrectly diagnoses3% of all adults over 40without diabetes as havingthe disease.

Summary

(b) Find the probability that a randomlyselected adult over 40 does not havediabetes, and is diagnosed as havingdiabetes.

“does not have diabetes” (A′)“is diagnosed as having diabetes”(D′)P(A′ ∩ D) = P(A′)P(D|A′)P(A′) =(0.95)(0.03) = .0285

Summary

“does not have diabetes” (A′)

“is diagnosed as having diabetes”(D′)P(A′ ∩ D) = P(A′)P(D|A′)P(A′) =(0.95)(0.03) = .0285

Summary

“does not have diabetes” (A′)“is diagnosed as having diabetes”(D′)

P(A′ ∩ D) = P(A′)P(D|A′)P(A′) =(0.95)(0.03) = .0285

.03 D.03

Summary

“does not have diabetes” (A′)“is diagnosed as having diabetes”(D′)P(A′ ∩ D) = P(A′)P(D|A′)P(A′) =(0.95)(0.03) = .0285

.03 D.03

Summary

(c) Find the probability that a randomlyselected adult over 40 is diagnosed asnot having diabetes.

There are two cases. Such a personeither

does not have it and is (correctly)diagnosed as not having it,

P(A′ ∩ D′) = P(A′)P(D′|A′)

has it, but is (incorrectly) diagnosedas not having it,

P(A ∩ D′) = P(A)P(D′|A)

We are looking forP(D′).

.05 A.10

A′ .97D′

Summary

P(A′ ∩ D′) = P(A′)P(D′|A′)

P(A ∩ D′) = P(A)P(D′|A)

.05 A.10

A′ .97D′

Summary

P(A′ ∩ D′) = P(A′)P(D′|A′)

P(A ∩ D′) = P(A)P(D′|A)

.05 A.10

A′ .97D′

Summary

P(A′ ∩ D′) = P(A′)P(D′|A′)

P(A ∩ D′) = P(A)P(D′|A)

.05 A.10

A′ .97D′

Summary

P(D′) = P(A′ ∩ D′) + P(A ∩ D′)

= P(A′)P(D′|A′) + P(A)P(D′|A)= (0.95)(0.97) + (0.05)(0.1)= 0.9215 + 0.005 = 0.9265

So, 92.65% of adults over 40 who take this test will bediagnosed as not having diabetes.

Summary

P(D′) = P(A′ ∩ D′) + P(A ∩ D′)

= P(A′)P(D′|A′) + P(A)P(D′|A)

= (0.95)(0.97) + (0.05)(0.1)= 0.9215 + 0.005 = 0.9265

Summary

P(D′) = P(A′ ∩ D′) + P(A ∩ D′)

= P(A′)P(D′|A′) + P(A)P(D′|A)= (0.95)(0.97) + (0.05)(0.1)

= 0.9215 + 0.005 = 0.9265

Summary

P(D′) = P(A′ ∩ D′) + P(A ∩ D′)

= P(A′)P(D′|A′) + P(A)P(D′|A)= (0.95)(0.97) + (0.05)(0.1)= 0.9215 + 0.005 = 0.9265

Summary

P(D′) = P(A′ ∩ D′) + P(A ∩ D′)

= P(A′)P(D′|A′) + P(A)P(D′|A)= (0.95)(0.97) + (0.05)(0.1)= 0.9215 + 0.005 = 0.9265

Summary

(d) Find the probability that a randomly selected adult over 40actually has diabetes, given that he/she is diagnosed as nothaving diabetes.

Here we are looking for P(A|D′) =P(A ∩ D′)

P(D′).

From the previous part,

P(A ∩ D′) = P(A)P(D′|A) = (0.05)(0.1) = 0.005

And P(D′) = 0.9265. So,

P(A|D′) =P(A ∩ D′)

P(D′)=

0.0050.9265

≈ 0.0054

So 0.54% of adults diagnosed as not having diabetes by thisdoctor actually do have the disease.

Summary

P(D′).

P(A ∩ D′) = P(A)P(D′|A) = (0.05)(0.1) = 0.005

And P(D′) = 0.9265. So,

P(A|D′) =P(A ∩ D′)

P(D′)=

0.0050.9265

≈ 0.0054

Summary

P(D′).

P(A ∩ D′) = P(A)P(D′|A) = (0.05)(0.1) = 0.005

And P(D′) = 0.9265. So,

P(A|D′) =P(A ∩ D′)

P(D′)=

0.0050.9265

≈ 0.0054

Summary

P(D′).

P(A ∩ D′) = P(A)P(D′|A) = (0.05)(0.1) = 0.005

And P(D′) = 0.9265. So,

P(A|D′) =P(A ∩ D′)

P(D′)=

0.0050.9265

≈ 0.0054

Summary

P(D′).

P(A ∩ D′) = P(A)P(D′|A) = (0.05)(0.1) = 0.005

And P(D′) = 0.9265. So,

P(A|D′) =P(A ∩ D′)

P(D′)=

0.0050.9265

≈ 0.0054

Summary

P(D′).

P(A ∩ D′) = P(A)P(D′|A) = (0.05)(0.1) = 0.005

And P(D′) = 0.9265. So,

P(A|D′) =P(A ∩ D′)

P(D′)=

0.0050.9265

≈ 0.0054

Summary

Independent Events

In probability, to say that two events are independentintuitively means that the occurence of one event makes itneither more nor less probable that the other occurs.

For example, if two cards are drawn with replacement froma deck of cards, the event of drawing a red card on the firsttrial and that of drawing a red card on the second trial areindependent.By contrast, if two cards are drawn without replacementfrom a deck of cards, the event of drawing a red card onthe first trial and that of drawing a red card on the secondtrial are not independent.

Summary

Independent Events

In probability, to say that two events are independentintuitively means that the occurence of one event makes itneither more nor less probable that the other occurs.For example, if two cards are drawn with replacement froma deck of cards, the event of drawing a red card on the firsttrial and that of drawing a red card on the second trial areindependent.

By contrast, if two cards are drawn without replacementfrom a deck of cards, the event of drawing a red card onthe first trial and that of drawing a red card on the secondtrial are not independent.

Summary

Independent Events

In probability, to say that two events are independentintuitively means that the occurence of one event makes itneither more nor less probable that the other occurs.For example, if two cards are drawn with replacement froma deck of cards, the event of drawing a red card on the firsttrial and that of drawing a red card on the second trial areindependent.By contrast, if two cards are drawn without replacementfrom a deck of cards, the event of drawing a red card onthe first trial and that of drawing a red card on the secondtrial are not independent.

Summary

Independent Events

Caution: very often intuition is not a reliable guide to thenotion of independence. Must use mathematics to verifyindependence, not intuition.

In problems we will either

be asked to determine whether or not two events areindependent, orbe asked to assume that two events are independent and touse that assumption to solve the problem.

Summary

Independent Events

Caution: very often intuition is not a reliable guide to thenotion of independence. Must use mathematics to verifyindependence, not intuition.In problems we will either

Summary

Independent Events

be asked to determine whether or not two events areindependent, or

be asked to assume that two events are independent and touse that assumption to solve the problem.

Summary

Independent Events

Summary

Definition (Independence)If A and B are any events in a sample space S, we say that Aand B are independent if and only if

P(A ∩ B) = P(A)P(B)

Otherwise, A and B are said to be dependent.

Summary

Example: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent?

We are asked to determine whether or not two events areindependent. To do this, we must determine whether or not

P(A ∩ B) = P(A)P(B)

We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B). This is aperfect place to use the addition principle:

P(A ∩ B) = P(A) + P(B)− P(A ∪ B) = 0.8 + 0.5− 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4

Since P(A ∩ B) = P(A)P(B) we conclude that A and B areindependent events.

Summary

P(A ∩ B) = P(A)P(B)

P(A ∩ B) = P(A) + P(B)− P(A ∪ B) = 0.8 + 0.5− 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4

Summary

P(A ∩ B) = P(A)P(B)

We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B).

This is aperfect place to use the addition principle:

P(A ∩ B) = P(A) + P(B)− P(A ∪ B) = 0.8 + 0.5− 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4

Summary

P(A ∩ B) = P(A)P(B)

P(A ∩ B) = P(A) + P(B)− P(A ∪ B) = 0.8 + 0.5− 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4

Summary

P(A ∩ B) = P(A)P(B)

P(A ∩ B) = P(A) + P(B)− P(A ∪ B) = 0.8 + 0.5− 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4

Summary

P(A ∩ B) = P(A)P(B)

P(A ∩ B) = P(A) + P(B)− P(A ∪ B) = 0.8 + 0.5− 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4

Summary

P(A ∩ B) = P(A)P(B)

P(A ∩ B) = P(A) + P(B)− P(A ∪ B) = 0.8 + 0.5− 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4

Summary

Example: On a stormy night, the probability that the electricitygoes out is 7% and the probability that the phone goes out is3%. Assuming that the two are independent, what is theprobability that neither the electricity nor the phone goes out ona stormy night?

Here we are asked to assume that two events are independent,and we will use this assumption to solve the problem

Let E represent the electricity going out. So P(E) = 0.07.Let F represent the phone going out. So P(F ) = 0.03.The probability that neither the electricity nor the phonegoes out is given by

P(E ′ ∩ F ′)

Summary

P(E ′ ∩ F ′)

Summary

Let E represent the electricity going out. So P(E) = 0.07.

Let F represent the phone going out. So P(F ) = 0.03.The probability that neither the electricity nor the phonegoes out is given by

P(E ′ ∩ F ′)

Summary

Let E represent the electricity going out. So P(E) = 0.07.Let F represent the phone going out. So P(F ) = 0.03.

The probability that neither the electricity nor the phonegoes out is given by

P(E ′ ∩ F ′)

Summary

P(E ′ ∩ F ′)

Summary

If E and F are independent, so are E ′ and F ′. So

P(E ′ ∩ F ′) = P(E ′)P(F ′)

Then to solve the problem:

P(E ′ ∩ F ′) = P(E ′)P(F ′)

= (1− P(E))(1− P(F ))

= (1− 0.07)(1− 0.03) = (0.93)(0.97)= 0.9021

So, the probability that neither the electricity nor the phonegoes out is 90.21%.

Summary

P(E ′ ∩ F ′) = P(E ′)P(F ′)

= (1− P(E))(1− P(F ))

= (1− 0.07)(1− 0.03) = (0.93)(0.97)= 0.9021

Summary

P(E ′ ∩ F ′) = P(E ′)P(F ′)

= (1− P(E))(1− P(F ))

= (1− 0.07)(1− 0.03) = (0.93)(0.97)= 0.9021

Summary

P(E ′ ∩ F ′) = P(E ′)P(F ′)

= (1− P(E))(1− P(F ))

= (1− 0.07)(1− 0.03) = (0.93)(0.97)

= 0.9021

Summary

P(E ′ ∩ F ′) = P(E ′)P(F ′)

= (1− P(E))(1− P(F ))

= (1− 0.07)(1− 0.03) = (0.93)(0.97)= 0.9021

Summary

P(E ′ ∩ F ′) = P(E ′)P(F ′)

= (1− P(E))(1− P(F ))

= (1− 0.07)(1− 0.03) = (0.93)(0.97)= 0.9021

Summary

TheoremIf A and B are independent events with nonzero probabilities ina sample space S, then

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

If either equation holds, then A and B are independent.

Summary

Example: A card is drawn from a standard 52 card deck.Events M and N are

M = the drawn card is a diamond (♦)N = the drawn card is even (face cards are not valued)

(a) Find P(N|M).

Notice that N ∩M is the set of all even diamonds. This is,

N ∩M = {2♦,4♦,6♦,8♦,10♦}

So n(N ∩M) = 5.

Summary

(a) Find P(N|M).

N ∩M = {2♦,4♦,6♦,8♦,10♦}

So n(N ∩M) = 5.

Summary

(a) Find P(N|M).

N ∩M = {2♦,4♦,6♦,8♦,10♦}

So n(N ∩M) = 5.

Summary

(a) Find P(N|M).

N ∩M = {2♦,4♦,6♦,8♦,10♦}

So n(N ∩M) = 5.

Summary

M is the set of all diamonds, so n(M) = 13.

P(N|M) =n(N ∩M)

(b) Test M and N for independence.

Here we could check whether P(M ∩ N) = P(M)P(N), but it isslightly more convenient to check whether P(N|M) = P(N).

N is the set of all even cards in the deck. There are 5 evencards per suit and 4 suits, so n(N) = 20. Then

P(N) =2052

P(N|M) = P(N) so yes, N and M are independent.

Summary

P(N|M) =n(N ∩M)

P(N) =2052

Summary

P(N|M) =n(N ∩M)

P(N) =2052

Summary

P(N|M) =n(N ∩M)

P(N) =2052

Summary

P(N|M) =n(N ∩M)

P(N) =2052

Summary

P(N|M) =n(N ∩M)

N is the set of all even cards in the deck. There are 5 evencards per suit and 4 suits, so n(N) = 20.

P(N) =2052

Summary

P(N|M) =n(N ∩M)

P(N) =2052

Summary

P(N|M) =n(N ∩M)

P(N) =2052

Summary

Definition (Independent Set of Events)A set of events is said to be independent if for each finitesubset {E1,E2, . . . ,Ek}

P(E1 ∩ E2 ∩ · · · ∩ Ek ) = P(E1)P(E2) · · ·P(Ek )

Summary

Example: One interpretation of a baseball player’s battingaverage is as the probability of getting a hit each time the playergoes to bat.

For instance, a player with a .300 average has probability .3 ofgetting a hit.

If a player with a .300 batting average bats four times in a gameand each at-bat is an independent event, what is the probabilityof the player getting at least one hit in the game?

Summary

Let Hi be the probability that the player gets a hit at his i th timeat bat. Then

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4) = 1− P(H ′1 ∩ H ′

2 ∩ H ′3 ∩ H ′

= 1− P(H ′1)P(H ′

2)P(H ′3)P(H ′

= 1− (0.7)4 = 0.7599

So if a player with a batting average of .300 bats four times in agame, then there is about a 76% chance of that player getting ahit.

Summary

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4) = 1− P(H ′1 ∩ H ′

2 ∩ H ′3 ∩ H ′

= 1− P(H ′1)P(H ′

2)P(H ′3)P(H ′

= 1− (0.7)4 = 0.7599

Summary

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4) = 1− P(H ′1 ∩ H ′

2 ∩ H ′3 ∩ H ′

= 1− P(H ′1)P(H ′

2)P(H ′3)P(H ′

= 1− (0.7)4 = 0.7599

Summary

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4) = 1− P(H ′1 ∩ H ′

2 ∩ H ′3 ∩ H ′

= 1− P(H ′1)P(H ′

2)P(H ′3)P(H ′

= 1− (0.7)4 = 0.7599

Summary

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4) = 1− P(H ′1 ∩ H ′

2 ∩ H ′3 ∩ H ′

= 1− P(H ′1)P(H ′

2)P(H ′3)P(H ′

= 1− (0.7)4 = 0.7599

Summary

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4) = 1− P(H ′1 ∩ H ′

2 ∩ H ′3 ∩ H ′

= 1− P(H ′1)P(H ′

2)P(H ′3)P(H ′

= 1− (0.7)4 = 0.7599

Summary

P(A|B) =P(A ∩ B)

P(B)P(B|A) = P(B ∩ A)

Note: P(A|B) is a probability based on the new sample spaceB, while P(A ∩ B) is based on the original sample space S.

Product Rule

P(A ∩ B) = P(B|A)P(A) = P(A|B)P(B)

Summary

Independent EventsA and B are independent if and only if

P(A ∩ B) = P(A)P(B)

If A and B are independent events with nonzeroprobabilities, then

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

If A and B are independent events with nonzeroprobabilities and either P(A|B) = P(A) or P(B|A) = P(B),then A and B are independent.If E1,E2, . . . ,En are independent, then

P(E1 ∩ E2 ∩ . . . ∩ En) = P(E1)P(E2) · · ·P(En)

math 1300: section 8-3 conditional probability, intersection, and independence

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