One of the two events must occur. If the chan | vedclass

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(D) Let $p$ be the probability of the other event. Then the probability of the first event is $\frac{2}{3}p$.Since one of the two events must occur an

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$3:2$

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(D) Let $p$ be the probability of the other event. Then the probability of the first event is $\frac{2}{3}p$.Since one of the two events must occur and they are mutually exclusive,the sum of their probabilities must be $1$.Therefore,$p + \frac{2}{3}p = 1$.Solving for $p$: $\frac{5}{3}p = 1 \Rightarrow p = \frac{3}{5}$.The probability of the first event is $1 - \frac{3}{5} = \frac{2}{5}$.The odds in favour of the other event are given by the ratio of its probability to the probability of its complement (the first event).Odds in favour = $p : (1 - p) = \frac{3}{5} : \frac{2}{5} = 3 : 2$.

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other,then the odds in favour of the other are

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