P, Q and R try to hit the same target one aft | vedclass

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Question

(A) Let $P, Q, R$ be the events that $P, Q, R$ hit the target respectively.Given probabilities are $P(P) = \frac{2}{3}, P(Q) = \frac{3}{5}, P(R) = \fr

Vedclass · Accepted Answer

$\frac{26}{105}$

Vedclass · Answer

(A) Let $P, Q, R$ be the events that $P, Q, R$ hit the target respectively.Given probabilities are $P(P) = \frac{2}{3}, P(Q) = \frac{3}{5}, P(R) = \frac{5}{7}$.The probabilities of not hitting the target are $P(P') = 1 - \frac{2}{3} = \frac{1}{3}$,$P(Q') = 1 - \frac{3}{5} = \frac{2}{5}$,and $P(R') = 1 - \frac{5}{7} = \frac{2}{7}$.We need the probability that the target is hit by $P$ or $Q$ but not by $R$. This can happen in three mutually exclusive ways:$1$. $P$ hits,$Q$ misses,$R$ misses: $P(P) 	imes P(Q') 	imes P(R') = \frac{2}{3} 	imes \frac{2}{5} 	imes \frac{2}{7} = \frac{8}{105}$.$2$. $P$ misses,$Q$ hits,$R$ misses: $P(P') 	imes P(Q) 	imes P(R') = \frac{1}{3} 	imes \frac{3}{5} 	imes \frac{2}{7} = \frac{6}{105}$.$3$. $P$ hits,$Q$ hits,$R$ misses: $P(P) 	imes P(Q) 	imes P(R') = \frac{2}{3} 	imes \frac{3}{5} 	imes \frac{2}{7} = \frac{12}{105}$.Summing these probabilities: $\frac{8}{105} + \frac{6}{105} + \frac{12}{105} = \frac{26}{105}$.

$P, Q$ and $R$ try to hit the same target one after the other. If their probabilities of hitting the target are $\frac{2}{3}, \frac{3}{5}, \frac{5}{7}$ respectively,then the probability that the target is hit by $P$ or $Q$ but not by $R$ is

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