# Chapter 11 - Counting Methods and Probability Theory - 11.6 Events Involving Not and Or; Odds - Exercise Set 11.6: 65

a.) 9:91 b.) 91:9

#### Work Step by Step

a.) 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 find the odds in favor of a child in a one-parent household having a parent who is a college graduate. E: {9} P(E) = $\frac{9}{100}$ P(not E) = 1 - P(E) = 1 - $\frac{9}{100}$ =$\frac{100 - 9}{100}$ = $\frac{91}{100}$ Odds in Favor = $\frac{\frac{9}{100}}{\frac{91}{100}}$ = $\frac{9}{91}$ b.)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 = $\frac{\frac{91}{100}}{\frac{9}{100}}$ = $\frac{91}{9}$ The odds against E can also be found by reversing the ratio representing the odds in favor of E.

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