Answer
False.
Work Step by Step
The odds for $E$: $\frac{P(E)}{P(\text{not E})}$. The probability of $E$: $\frac{P(E)}{P(\text{not E})+P(E)}$
Thus the statement is false, because the probability is: $\frac{1}{5+1}=\frac{1}{6}$
Thinking Mathematically (6th Edition)
by Blitzer, Robert F.
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 - Concept and Vocabulary Check - Page 733: 9
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