   Chapter 9.3, Problem 18ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Just as the difference rule gives rise to a formula for the probability of the complement of an event, so the addition and inclusion/exclusion rules give rise to formulas for the probability of the union of mutually disjoint events and for a general union of (not necessarily mutually exclusive) events. Prove that for mutually disjoint events A and B,  P ( A ∪ B ) = P ( A ) + P ( B ) . Prove that for any events A and B, P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) .

To determine

(a)

To prove that for mutually disjoint events A and B ,

P(AB)=P(A)+P(B).

Explanation

Given information:

P(AB)

Concept used:

N(AB)=N(A)+N(B) for mutually disjoint sets.

Calculation:

The objective is to prove that P(AB)=P(A)+P(B) for mutually disjoint events A and B.

Let N(S) be the number of elements in the sample space S.

By addition rule, N(AB)=N(A)+N(B) for mutually disjoint sets.

Divide N(AB)=N(A)+N(B) by N(S).

N( AB)N(S)=N(A)+N(B)N(S)N( AB)N(S)=N(A)N(S)+N(B<

To determine

(b)

To prove that for any events A and B.

P(AB)=P(A)+P(B)P(AB)

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