The odds against an event are 5 : 2 and the o | vedclass

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(B) Let $A$ and $B$ be the two independent events.For event $A$,the odds against are $5 : 2$,so the probability of $A$ occurring is $P(A) = \frac{2}{5

Vedclass · Accepted Answer

$\frac{52}{77}$

Vedclass · Answer

(B) Let $A$ and $B$ be the two independent events.For event $A$,the odds against are $5 : 2$,so the probability of $A$ occurring is $P(A) = \frac{2}{5+2} = \frac{2}{7}$.The probability of $A$ not occurring is $P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.For event $B$,the odds in favor are $6 : 5$,so the probability of $B$ occurring is $P(B) = \frac{6}{6+5} = \frac{6}{11}$.The probability of $B$ not occurring is $P(B') = 1 - \frac{6}{11} = \frac{5}{11}$.The probability that at least one event occurs is $1 - P(A' \cap B')$.Since $A$ and $B$ are independent,$P(A' \cap B') = P(A') 	imes P(B') = \frac{5}{7} 	imes \frac{5}{11} = \frac{25}{77}$.Therefore,the required probability is $1 - \frac{25}{77} = \frac{77-25}{77} = \frac{52}{77}$.

The odds against an event are $5 : 2$ and the odds in favor of another independent event are $6 : 5$. What is the probability that at least one of the events occurs?

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