Math 1300: Section 8 -2 Union, Intersection, and Complement of Events; Odds

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Union and IntersectionComplement of an Event

OddsApplications to Empirical Probability

Math 1300 Finite MathematicsSection 8-2: Union, Intersection, and Complement of Events;

Odds

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics




In this section, we will develop the rules of probability forcompound events (more than one simple event) and willdiscuss probabilities involving the union of events as well asintersection of two events.





If A and B are two events in a sample space S, then the unionof A and B, denoted by A ∪ B, and the intersection of A and B,denoted by A ∩ B, are defined as follows:

Definition (Union: A ∪ B)

A B

A ∪ B = {e ∈ S|e ∈ A or e ∈ B}

The event “A or B” is defined as the set A ∪ B.





If A and B are two events in a sample space S, then the unionof A and B, denoted by A ∪ B, and the intersection of A and B,denoted by A ∩ B, are defined as follows:

Definition (Union: A ∪ B)

A B

A ∪ B = {e ∈ S|e ∈ A or e ∈ B}

The event “A or B” is defined as the set A ∪ B.Jason Aubrey Math 1300 Finite Mathematics




Definition (Intersection: A ∩ B)

A B

A ∩ B = {e ∈ S|e ∈ A and e ∈ B}

The event “A and B” is defined as the set A ∩ B.





Definition (Intersection: A ∩ B)

A B

A ∩ B = {e ∈ S|e ∈ A and e ∈ B}

The event “A and B” is defined as the set A ∩ B.





Example: Consider the sample space of equally likely eventsfor the rolling of a single fair die: S = {1,2,3,4,5,6}

(a) What is the probability of rolling an odd number and a primenumber?

Let A be the event “an odd number is rolled”. Let B be the event“a prime number is rolled”.

Since only the outcomes 3 and 5 are both odd and prime,A ∩ B = {3,5}.By the equally likely assumption,

P(A ∩ B) =n(A ∩ B)

n(S)=

26=

13









P(A ∩ B) =n(A ∩ B)

n(S)=

26=

13









P(A ∩ B) =n(A ∩ B)

n(S)=

26=

13








Since only the outcomes 3 and 5 are both odd and prime,A ∩ B = {3,5}.

By the equally likely assumption,

P(A ∩ B) =n(A ∩ B)

n(S)=

26=

13









P(A ∩ B) =n(A ∩ B)

n(S)=

26=

13





(b) What is the probability of rolling an odd number or a primenumber?

Again, A = {1,3,5} and B = {2,3,5} so

A ∪ B = {1,2,3,5}

By the equally likely assumption

P(A ∪ B) =n(A ∪ B)

n(S)=

46=

23






Again, A = {1,3,5} and B = {2,3,5} so

A ∪ B = {1,2,3,5}


P(A ∪ B) =n(A ∪ B)

n(S)=

46=

23






Again, A = {1,3,5} and B = {2,3,5} so

A ∪ B = {1,2,3,5}


P(A ∪ B) =n(A ∪ B)

n(S)=

46=

23






Again, A = {1,3,5} and B = {2,3,5} so

A ∪ B = {1,2,3,5}


P(A ∪ B) =n(A ∪ B)

n(S)=

46=

23





Suppose that an event E is

E = A ∪ B.

Is P(E) = P(A) + P(B)?Only if A and B are mutually exclusive (disjoint), that is,if A ∩ B = ∅.In this case, P(A ∪ B) is the sum of the probabilities of allof the simple events in A plus the sum of the probabilitiesof all of the simple events in B.






E = A ∪ B.

Is P(E) = P(A) + P(B)?

Only if A and B are mutually exclusive (disjoint), that is,if A ∩ B = ∅.In this case, P(A ∪ B) is the sum of the probabilities of allof the simple events in A plus the sum of the probabilitiesof all of the simple events in B.






E = A ∪ B.

Is P(E) = P(A) + P(B)?Only if A and B are mutually exclusive (disjoint), that is,if A ∩ B = ∅.

In this case, P(A ∪ B) is the sum of the probabilities of allof the simple events in A plus the sum of the probabilitiesof all of the simple events in B.






E = A ∪ B.

Is P(E) = P(A) + P(B)?Only if A and B are mutually exclusive (disjoint), that is,if A ∩ B = ∅.In this case, P(A ∪ B) is the sum of the probabilities of allof the simple events in A plus the sum of the probabilitiesof all of the simple events in B.





But what happens if A and B are not mutually exclusive;that is, what if A ∩ B 6= ∅?

In this case, we must use a version of the addition principlefor probability.





But what happens if A and B are not mutually exclusive;that is, what if A ∩ B 6= ∅?In this case, we must use a version of the addition principlefor probability.





Definition (Probability of a Union of Two Events)For any events A and B,

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

If A and B are mutually exclusive, then

P(A ∪ B) = P(A) + P(B)





Example Let E and F be events in a sample space S. IfP(E) = 0.35, P(F ) = 0.25 and P(E ∪ F ) = 0.55 then what isP(E ∩ F )?

Notice here that P(E ∪ F ) 6= P(E) + P(F ), so we know that theevents E and F are not mutually exclusive; that is,P(E ∩ F ) 6= 0.

To find P(E ∩ F ) we use the addition principle:

P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

0.55 = 0.35 + 0.25− P(E ∩ F )

P(E ∩ F ) = 0.35 + 0.25− 0.55 = 0.05








P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

0.55 = 0.35 + 0.25− P(E ∩ F )

P(E ∩ F ) = 0.35 + 0.25− 0.55 = 0.05








P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

0.55 = 0.35 + 0.25− P(E ∩ F )

P(E ∩ F ) = 0.35 + 0.25− 0.55 = 0.05








P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

0.55 = 0.35 + 0.25− P(E ∩ F )

P(E ∩ F ) = 0.35 + 0.25− 0.55 = 0.05








P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

0.55 = 0.35 + 0.25− P(E ∩ F )

P(E ∩ F ) = 0.35 + 0.25− 0.55 = 0.05





Example: A single card is drawn from a deck of cards. Find theprobability that the card is a jack or club.

Let E = “the set of jacks”, and F = “the set of clubs”. Are Eand F mutually exclusive? No because a card can be both ajack and a club?

Now, n(E) = 4 - there are four jacks; by the equally likelyassumption,

P(E) =n(E)

n(S)=

452

.

Also, n(F ) = 13 - there are thirteen clubs; by the equallylikely assumption,

P(F ) =n(F )

n(S)=

1352

.






Let E = “the set of jacks”, and F = “the set of clubs”. Are Eand F mutually exclusive?

No because a card can be both ajack and a club?


P(E) =n(E)

n(S)=

452

.


P(F ) =n(F )

n(S)=

1352

.








P(E) =n(E)

n(S)=

452

.


P(F ) =n(F )

n(S)=

1352

.








P(E) =n(E)

n(S)=

452

.


P(F ) =n(F )

n(S)=

1352

.








P(E) =n(E)

n(S)=

452

.


P(F ) =n(F )

n(S)=

1352

.





Finally, n(E ∩ F ) = 1 - there is one jack that is also a club;by the equally likely assumption,

P(E ∩ F ) =1

52.

Now we can use the addition principle to get

P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

P(E ∪ F ) =4

52+

1352− 1

52

=1652

=4

13






P(E ∩ F ) =1

52.


P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

P(E ∪ F ) =4

52+

1352− 1

52

=1652

=4

13






P(E ∩ F ) =1

52.


P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

P(E ∪ F ) =4

52+

1352− 1

52

=1652

=4

13






P(E ∩ F ) =1

52.


P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

P(E ∪ F ) =4

52+

1352− 1

52

=1652

=4

13





Example: Three coins are tossed. Assume they are fair coins.(Tossing three coins is the same experiment as tossing onecoin three times.)

(a) Use the multiplication principle to calculate the total numberof outcomes in the sample space.

n(S) = (2)(2)(2) = 8

(b) Find the probability of flipping at least two tails.

Let E be the event of flipping at least two tails.Let A be the event that exactly two tails are flipped.Let B be the event that exactly three tails are flipped.







n(S) = (2)(2)(2) = 8









n(S) = (2)(2)(2) = 8









n(S) = (2)(2)(2) = 8









n(S) = (2)(2)(2) = 8


Let E be the event of flipping at least two tails.

Let A be the event that exactly two tails are flipped.Let B be the event that exactly three tails are flipped.







n(S) = (2)(2)(2) = 8


Let E be the event of flipping at least two tails.Let A be the event that exactly two tails are flipped.

Let B be the event that exactly three tails are flipped.







n(S) = (2)(2)(2) = 8







Notice that E = A ∪ B and that A and B are mutuallyexclusive; that is,

E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)

P(A) =n({HTT ,THT ,TTH})

n(S)=

38

P(B) =n({TTT})

n(S)=

18

Therefore,

P(E) = P(A) + P(B) =38+

18=

12






E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)


n(S)=

38

P(B) =n({TTT})

n(S)=

18

Therefore,

P(E) = P(A) + P(B) =38+

18=

12






E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)


n(S)=

38

P(B) =n({TTT})

n(S)=

18

Therefore,

P(E) = P(A) + P(B) =38+

18=

12






E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)


n(S)=

38

P(B) =n({TTT})

n(S)=

18

Therefore,

P(E) = P(A) + P(B) =38+

18=

12






E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)


n(S)=

38

P(B) =n({TTT})

n(S)=

18

Therefore,

P(E) = P(A) + P(B) =38+

18=

12





Suppose that we divide a finite sample space

S = {e1, . . . ,en}

into two subsets E and E ′ such that

E ∩ E ′ = ∅.

That is, E and E ′ are mutually exclusive, and

E ∪ E ′ = S.

Then E ′ is called the complement of E relative to S. The setE ′ contains all the elements of S that are not in E .






S = {e1, . . . ,en}


E ∩ E ′ = ∅.


E ∪ E ′ = S.







S = {e1, . . . ,en}


E ∩ E ′ = ∅.


E ∪ E ′ = S.

Then E ′ is called the complement of E relative to S.

The setE ′ contains all the elements of S that are not in E .






S = {e1, . . . ,en}


E ∩ E ′ = ∅.


E ∪ E ′ = S.






Furthermore,

P(S) = P(E ∪ E ′)

= P(E) + P(E ′) = 1

Therefore,

P(E) = 1− P(E ′) P(E ′) = 1− P(E)

Many times it is easier to first compute the probability that andevent won’t occur, and then use that to find the probability thatthe event will occur.





Furthermore,

P(S) = P(E ∪ E ′)

= P(E) + P(E ′) = 1

Therefore,

P(E) = 1− P(E ′) P(E ′) = 1− P(E)






Furthermore,

P(S) = P(E ∪ E ′)

= P(E) + P(E ′) = 1

Therefore,

P(E) = 1− P(E ′) P(E ′) = 1− P(E)






Example: If A and B are mutually exclusive events withP(A) = 0.6 and P(B) = 0.3, find the following probabilities:

(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)

Since A ∩ B = ∅, A ⊆ B′ so A ∩ B′ = A and

P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)

P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)

P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6






(a) P(A ∩ B)P(A ∩ B) = 0

(b) P(A ∩ B)

P(A ∩ B) = P(A) + P(B)− P(A ∩ B) = 0.6 + 0.3− 0 = 0.9

(c) P(A′)P(A′) = 1− P(A) = 1− 0.6 = 0.4

(d) P(A ∩ B′)


P(A ∩ B′) = P(A) = 0.6





Example: What is the probability that when two dice aretossed, the number of points on each die will not be the same?

This is the same as saying that doubles will not occur. Forexample,

E be the set of all rolls of two dice which do not result indoubles. Mathematically we can represent this as

E = {(n,m)|1 ≤ n,m ≤ 6 and n 6= m}We wish to find P(E). Let S be be the sample space for thisexperiment.

S = {(n,m)|1 ≤ n,m ≤ 6} and n(S) = 36









S = {(n,m)|1 ≤ n,m ≤ 6} and n(S) = 36








E = {(n,m)|1 ≤ n,m ≤ 6 and n 6= m}We wish to find P(E).

Let S be be the sample space for thisexperiment.

S = {(n,m)|1 ≤ n,m ≤ 6} and n(S) = 36









S = {(n,m)|1 ≤ n,m ≤ 6} and n(S) = 36





Here we have

E ′ = {(n,m)|1 ≤ n,m ≤ 6 and n = m}= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

Since

P(E ′) =n(E ′)

n(S)=

636

=16

we haveP(E) = 1− P(E ′) = 1− 1

6=

56





Here we have

E ′ = {(n,m)|1 ≤ n,m ≤ 6 and n = m}

= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

Since

P(E ′) =n(E ′)

n(S)=

636

=16

we haveP(E) = 1− P(E ′) = 1− 1

6=

56





Here we have

E ′ = {(n,m)|1 ≤ n,m ≤ 6 and n = m}= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

Since

P(E ′) =n(E ′)

n(S)=

636

=16

we haveP(E) = 1− P(E ′) = 1− 1

6=

56





Here we have

E ′ = {(n,m)|1 ≤ n,m ≤ 6 and n = m}= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

Since

P(E ′) =n(E ′)

n(S)=

636

=16

we haveP(E) = 1− P(E ′) = 1− 1

6=

56





Here we have

E ′ = {(n,m)|1 ≤ n,m ≤ 6 and n = m}= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

Since

P(E ′) =n(E ′)

n(S)=

636

=16

we haveP(E) = 1− P(E ′) = 1− 1

6=

56





Example: A coin is tossed 5 times. What is the probability thatheads turn up at least once?

Let E represent the event “heads turns up at least once” and letS represent the sample space. Notice first that

S = {HTTHT ,HHHTT ,THHHT ,HTHTH,TTTTT ,HHHHT , . . .}

We assume the coin is fair so that we may also assume that allof the outcomes in the sample space are equally likely. What isn(S)?

n(S) = (2)(2)(2)(2)(2) = 32

E contains all outcomes that have at least one H. E.g. HTTHT ,HHHTT , etc.









n(S) = (2)(2)(2)(2)(2) = 32










n(S) = (2)(2)(2)(2)(2) = 32










n(S) = (2)(2)(2)(2)(2) = 32










n(S) = (2)(2)(2)(2)(2) = 32






What is in the set E ′?

The opposite of “heads turn up at leastonce” is “heads do not turn up at all.” So,

E ′ = {TTTTT} and P(E ′) =1

32

Therefore,

P(E) = 1− P(E ′) = 1− 132

=3132

Tip: Consider using complements whenever you encounter aprobability (or even counting problems) that contains the phrase“at least once”.





What is in the set E ′? The opposite of “heads turn up at leastonce” is “heads do not turn up at all.” So,


32

Therefore,

P(E) = 1− P(E ′) = 1− 132

=3132








32

Therefore,

P(E) = 1− P(E ′) = 1− 132

=3132








32

Therefore,

P(E) = 1− P(E ′) = 1− 132

=3132






Example: A shipment of 40 precision parts, including 8 that aredefective, is sent to an assembly plant. The quality controldivision selects 10 at random for testing and rejects theshipment if 1 or more in the sample are found defective. Whatis the probability that the shipment will be rejected?

Notice first that the question

What is the probability that the shipment will berejected?

is really asking

What is the probability that the 10 parts selected fortesting contain at least one defective part?





Example: A shipment of 40 precision parts, including 8 that aredefective, is sent to an assembly plant. The quality controldivision selects 10 at random for testing and rejects theshipment if 1 or more in the sample are found defective. Whatis the probability that the shipment will be rejected?

Notice first that the question

What is the probability that the shipment will berejected?

is really asking

What is the probability that the 10 parts selected fortesting contain at least one defective part?





Let S be all possible selections of 10 parts from the shipment of40.

n(S) = C(40,10) = 847,660,528

Let E be the set of all selections of 10 parts that contain at leastone defective part. We want to find P(E).

Notice that every set of 10 parts eithercontains at least one defective part (so is in E), orcontains no defective parts

Thus E ′ is the set of all selections of 10 parts that contain nodefective parts.






n(S) = C(40,10) = 847,660,528









n(S) = C(40,10) = 847,660,528









n(S) = C(40,10) = 847,660,528


Notice that every set of 10 parts either

contains at least one defective part (so is in E), orcontains no defective parts







n(S) = C(40,10) = 847,660,528


Notice that every set of 10 parts eithercontains at least one defective part (so is in E), or

contains no defective partsThus E ′ is the set of all selections of 10 parts that contain nodefective parts.






n(S) = C(40,10) = 847,660,528









n(S) = C(40,10) = 847,660,528








The shipment of 40 parts contains 8 that are defective. To pick10 that have no defective parts, we choose from the 32 that arenot defective, so

n(E ′) = C(32,10) = 64,512,240

Therefore,

P(E) = 1− P(E ′) = 1− n(E ′)

n(S)

= 1− 64,512,240847,660,528

≈ 1− 0.0761

≈ 0.9239

So there is about a 92.4% chance that the shipment will berejected.






n(E ′) = C(32,10) = 64,512,240

Therefore,

P(E) = 1− P(E ′) = 1− n(E ′)

n(S)

= 1− 64,512,240847,660,528

≈ 1− 0.0761

≈ 0.9239







n(E ′) = C(32,10) = 64,512,240

Therefore,

P(E) = 1− P(E ′) = 1− n(E ′)

n(S)

= 1− 64,512,240847,660,528

≈ 1− 0.0761

≈ 0.9239







n(E ′) = C(32,10) = 64,512,240

Therefore,

P(E) = 1− P(E ′) = 1− n(E ′)

n(S)

= 1− 64,512,240847,660,528

≈ 1− 0.0761

≈ 0.9239







n(E ′) = C(32,10) = 64,512,240

Therefore,

P(E) = 1− P(E ′) = 1− n(E ′)

n(S)

= 1− 64,512,240847,660,528

≈ 1− 0.0761

≈ 0.9239







n(E ′) = C(32,10) = 64,512,240

Therefore,

P(E) = 1− P(E ′) = 1− n(E ′)

n(S)

= 1− 64,512,240847,660,528

≈ 1− 0.0761

≈ 0.9239






Definition (From Probabilities to Odds)

If P(E) is the probability of the event E , then1 the odds for E are given by

P(E)

1− P(E)=

P(E)

P(E ′),P(E) 6= 1.

2 the odds against E are given by

1− P(E)

P(E)=

P(E ′)

P(E),P(E) 6= 0

Note: When possible, odds are to be expressed as ratios ofwhole numbers.







P(E)

1− P(E)=

P(E)

P(E ′),P(E) 6= 1.


1− P(E)

P(E)=

P(E ′)

P(E),P(E) 6= 0








P(E)

1− P(E)=

P(E)

P(E ′),P(E) 6= 1.


1− P(E)

P(E)=

P(E ′)

P(E),P(E) 6= 0








P(E)

1− P(E)=

P(E)

P(E ′),P(E) 6= 1.


1− P(E)

P(E)=

P(E ′)

P(E),P(E) 6= 0






Example: Given the following probabilities for an event E , findthe odds for and against E :

P(E) = 35

Since P(E) = 3/5 we know P(E ′) = 1− P(E) = 2/5.Then the odds for E are

P(E)

P(E ′)=

3/52/5

=32

And the odds against E are

P(E ′)

P(E)=

2/53/5

=23






P(E) = 35


P(E)

P(E ′)=

3/52/5

=32


P(E ′)

P(E)=

2/53/5

=23






P(E) = 35

Since P(E) = 3/5 we know P(E ′) = 1− P(E) = 2/5.

Then the odds for E are

P(E)

P(E ′)=

3/52/5

=32


P(E ′)

P(E)=

2/53/5

=23






P(E) = 35


P(E)

P(E ′)=

3/52/5

=32


P(E ′)

P(E)=

2/53/5

=23






P(E) = 35


P(E)

P(E ′)=

3/52/5

=32


P(E ′)

P(E)=

2/53/5

=23





P(E) = 0.35

Since P(E) = 0.35 we know P(E ′) = 1− 0.35 = 0.65Then the odds for E are

P(E)

P(E ′)=

0.350.65

=7

13


P(E ′)

P(E)=

0.650.35

=137





P(E) = 0.35Since P(E) = 0.35 we know P(E ′) = 1− 0.35 = 0.65

Then the odds for E are

P(E)

P(E ′)=

0.350.65

=7

13


P(E ′)

P(E)=

0.650.35

=137





P(E) = 0.35Since P(E) = 0.35 we know P(E ′) = 1− 0.35 = 0.65Then the odds for E are

P(E)

P(E ′)=

0.350.65

=7

13


P(E ′)

P(E)=

0.650.35

=137





P(E) = 0.35Since P(E) = 0.35 we know P(E ′) = 1− 0.35 = 0.65Then the odds for E are

P(E)

P(E ′)=

0.350.65

=7

13


P(E ′)

P(E)=

0.650.35

=137





Example: Find the odds in favor of rolling a total of seven whentwo dice are tossed.

Let E be the event that the sum of the two dice is seven. So,

E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

n(E) = 6, n(S) = 36,

P(E) =6

36and P(E ′) =

3036

ThereforeP(E)

P(E ′)=

6/3630/36

=6

30=

15







E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

n(E) = 6, n(S) = 36,

P(E) =6

36and P(E ′) =

3036

ThereforeP(E)

P(E ′)=

6/3630/36

=6

30=

15







E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

n(E) = 6, n(S) = 36,

P(E) =6

36and P(E ′) =

3036

ThereforeP(E)

P(E ′)=

6/3630/36

=6

30=

15







E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

n(E) = 6, n(S) = 36,

P(E) =6

36and P(E ′) =

3036

ThereforeP(E)

P(E ′)=

6/3630/36

=6

30=

15







E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

n(E) = 6, n(S) = 36,

P(E) =6

36and P(E ′) =

3036

ThereforeP(E)

P(E ′)=

6/3630/36

=6

30=

15





If the odds for an event E areab

, then the probability of Eis,

P(E) =a

a + b

If the odds against an event E areab

then the probability ofE is

P(E) =b

a + b





If the odds for an event E areab

, then the probability of Eis,

P(E) =a

a + b

If the odds against an event E areab

then the probability ofE is

P(E) =b

a + b





Example: If in repeated rolls of two fair dice the odds againstrolling a 6 before rolling a 7 are 6:5, what is the probability ofrolling a 6 before rolling a 7?

Let E be the event “a 6 is rolled before a 7 is rolled”.odds against E are 6:5Therefore,

P(E) =5

6 + 5=

511






Let E be the event “a 6 is rolled before a 7 is rolled”.

odds against E are 6:5Therefore,

P(E) =5

6 + 5=

511






Let E be the event “a 6 is rolled before a 7 is rolled”.odds against E are 6:5

Therefore,

P(E) =5

6 + 5=

511






Let E be the event “a 6 is rolled before a 7 is rolled”.odds against E are 6:5Therefore,

P(E) =5

6 + 5=

511





Example: The data below was obtained from a random surveyof 1,000 residents of a state. The participants were asked theirpolitical affiliations and their preferences in an upcominggubernatorial election (D = Democrat, R = Republican, U =Unaffiliated. )

D R U TotalsCandidate A 200 100 85 385Candidate B 250 230 50 530No Preference 50 20 15 85Totals 500 350 150 1,000

If a resident of the state is selected at random, what is theempirical probability that the resident is not affiliated with apolitical party or has no preference? What are the odds for thisevent?





Example: The data below was obtained from a random surveyof 1,000 residents of a state. The participants were asked theirpolitical affiliations and their preferences in an upcominggubernatorial election (D = Democrat, R = Republican, U =Unaffiliated. )


If a resident of the state is selected at random, what is theempirical probability that the resident is not affiliated with apolitical party or has no preference? What are the odds for thisevent?






We are looking for P(U ∪ N):

P(U ∪ N) = P(U) + P(N)− P(U ∩ N)

=150

1000+

851000

− 151000

=220

1000= 0.22 or 22%

Then the odds for this event are22

100− 22=

2278

=1139







P(U ∪ N) = P(U) + P(N)− P(U ∩ N)

=150

1000+

851000

− 151000

=220

1000= 0.22 or 22%


100− 22=

2278

=1139







P(U ∪ N) = P(U) + P(N)− P(U ∩ N)

=150

1000+

851000

− 151000

=220

1000= 0.22 or 22%


100− 22=

2278

=1139







P(U ∪ N) = P(U) + P(N)− P(U ∩ N)

=150

1000+

851000

− 151000

=220

1000= 0.22 or 22%


100− 22=

2278

=1139







P(U ∪ N) = P(U) + P(N)− P(U ∩ N)

=150

1000+

851000

− 151000

=220

1000= 0.22 or 22%


100− 22=

2278

=1139


Education

Math 1300: Section 8 -2 Union, Intersection, and Complement of Events; Odds