If the odds against A solving a problem are 4 | vedclass

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Question

(C) Let $P(A)$ be the probability that $A$ solves the problem and $P(B)$ be the probability that $B$ solves the problem.Given odds against $A$ are $4:

Vedclass · Accepted Answer

$43/84$

Vedclass · Answer

(C) Let $P(A)$ be the probability that $A$ solves the problem and $P(B)$ be the probability that $B$ solves the problem.Given odds against $A$ are $4:3$,so $P(A') = 4/7$ and $P(A) = 1 - 4/7 = 3/7$.Given odds in favor of $B$ are $7:5$,so $P(B) = 7/(7+5) = 7/12$ and $P(B') = 1 - 7/12 = 5/12$.The probability that only one of them solves the problem is $P(A \cap B') + P(A' \cap B)$.$P(A \cap B') = P(A) 	imes P(B') = (3/7) 	imes (5/12) = 15/84$.$P(A' \cap B) = P(A') 	imes P(B) = (4/7) 	imes (7/12) = 28/84$.Total probability = $15/84 + 28/84 = 43/84$.

If the odds against $A$ solving a problem are $4:3$ and the odds in favor of $B$ solving the problem are $7:5$,what is the probability that only one of them solves the problem?

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