## 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: 56

#### Answer

As a ratio-57:85; 85:57

#### Work Step by Step

The odds in favor of E are found by taking the probability that E will occur and dividing 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 Army. P(E)= $\frac{570000}{1420000}$ = $\frac{57}{142}$ P(not E) = 1 - P(E) = 1 - $\frac{57}{142}$ =$\frac{142 - 57}{142}$ = $\frac{85}{142}$ Odds in Favor = $\frac{\frac{57}{142}}{\frac{85}{142}}$ = $\frac{57}{85}$ 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 = $\frac{\frac{85}{142}}{\frac{57}{142}}$ = $\frac{85}{57}$ The odds against E can also be found by reversing the ratio representing the odds in favor of E.

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