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 - Page 736: 89

Answer

See below.

Work Step by Step

The odds for $E$: $\frac{P(E)}{P(\text{not E})}$. The probability for $E$: $\frac{P(E)}{P(\text{not E})+P(E)}$. The odds against $E$: $\frac{P(\text{not E})}{P(E)}$ The probability against $E$: $\frac{P(\text{not E})}{P(\text{not E})+P(E)}$. Hence here the probability for making it is: $\frac{21}{21+4}=\frac{21}{25}$, thus the probability of missing it: $1-\frac{21}{25}=\frac{4}{25}$. Out of $100$ shots on average he made: $100\cdot\frac{21}{25}=4\cdot21=84$.
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