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Chapter 3: Q. 3.26 (page 111)

show thatP(A鈭�B)=1, thenPBc鈭�Ac=1.

P(A鈭�B)=1, thenPBc鈭�Ac=1

Short Answer

Expert verified

The probability ofP(A鈭�B)=1鈬�B鈯�A鈬�Ac鈯�Bc鈬�PBc鈭�Ac=1

Step by step solution

01

Step1:P(A∣B)=1

P(A鈭�B)=1

Logical proof: If A is certain with the knowledge of B this is equivalent to logical formulation:

B鈫�A

There is a logical rule called modus toles that states:

B鈫�A鈬�Ac鈫�Bc

And, in the same way that we converted probability to logic:

Ac鈫�Bc鈬�PBc鈭�Ac=1

02

Step2:PBc∣Ac=1

P(A鈭�B)=1鈬�P(AB)P(B)=1鈬�P(AB)=P(B)

PAcBc=PAc鈭�PAcB

=PAc鈭�[P(B)鈭�P(AB)]

=PAc鈭�0

=PAc

And going in a different direction than before yields:

PAcBc=PBc鈬�PAcBcPBc=1鈬�PAc鈭�Bc=1

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Most popular questions from this chapter

In Example 5e, what is the conditional probability that the ith coin was selected given that the first n trials all result in heads?

Prove that if E1,E2,鈥�,Enare independent events, then

PE1鈭�E2鈭�鈰�鈭�En=1-鈭�i=1n1-PEi

What is the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice is i,i=2,3,...,12?

Consider a school community of mfamilies, with niof them having ichildren, i=1,鈥�,k,鈭�i=1kni=mConsider the following two methods for choosing a child:

1. Choose one of the mfamilies at random and then randomly choose a child from that family.

2. Choose one of the 鈭�i=1kinichildren at random.

Show that method 1is more likely than method 2to result

in the choice of a firstborn child.

Hint: In solving this problem, you will need to show that

鈭�i=1kini鈭�j=1knjj鈮�鈭�i=1kni鈭�j=1knj

To do so, multiply the sums and show that for all pairs i,j, the coefficient of the termninj is greater in the expression on the left than in the one on the right.

Repeat Problem 3.84 when each of the 3 players

selects from his own urn. That is, suppose that there are

3 different urns of 12 balls with 4 white balls in each urn.
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