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: $\frac{1}{5+1}=\frac{1}{6}$
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