## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 11 - Counting Methods and Probability Theory - 11.6 Events Involving Not and Or; Odds - Exercise Set 11.6 - Page 736: 54

#### Answer

$\frac{89}{142}$

#### Work Step by Step

The probability that an event E will not occur is equal to 1 minus the probability that it will occur. P(not E)= 1 - P(E) E: is in the Army or the Navy. A = Neither in Army nor in Navy P( is in the Army or the Navy) = 1 - P( Neither in Army nor in Navy) P( Neither in Army nor in Navy) = $\frac{530000}{1420000}$ = $\frac{53}{142}$ P( is in the Army or the Navy) = 1 - $\frac{53}{142}$ = $\frac{142 - 53}{142}$ = $\frac{89}{142}$

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