Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.6 Events Involving Not and Or; Odds - Exercise Set 11.6: 55

Answer

$\frac{55}{16}$ , $\frac{16}{55}$

Work Step by Step

The odds in favor of E are found by taking the probability that E will occur and dividing it by the probability that E will not occur. Odds in Favor = $\frac{P(E)}{P(not E)}$ We are asked to find the odds in favor and the odds against a person being in the Navy. P(E)= $\frac{320000}{1420000}$ = $\frac{32}{142}$ P(not E) = 1 - P(E) = 1 - $\frac{32}{142}$ =$\frac{142 - 32}{142}$ = $\frac{110}{142}$ = $\frac{55}{71}$ Odds in Favor = $\frac{\frac{16}{71}}{\frac{55}{71}}$ = $\frac{16}{55}$ The odds against E are found by taking the probability that E will not occur and dividing by the probability that E will occur. Odds against E = $\frac{P(not E)}{P(E)}$ Odds against E = $\frac{\frac{55}{71}}{\frac{16}{71}}$ = $\frac{55}{16}$
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