Three persons P, Q and R independently try to | vedclass

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(A) Let $P(P), P(Q), P(R)$ be the probabilities of hitting the target by $P, Q, R$ respectively. $P(P) = \frac{3}{4}, P(Q) = \frac{1}{2}, P(R) = \frac

Vedclass · Accepted Answer

$\frac{21}{64}$

Vedclass · Answer

(A) Let $P(P), P(Q), P(R)$ be the probabilities of hitting the target by $P, Q, R$ respectively. $P(P) = \frac{3}{4}, P(Q) = \frac{1}{2}, P(R) = \frac{5}{8}$. The probabilities of missing the target are $P(P') = 1 - \frac{3}{4} = \frac{1}{4}$,$P(Q') = 1 - \frac{1}{2} = \frac{1}{2}$,and $P(R') = 1 - \frac{5}{8} = \frac{3}{8}$. The event that the target is hit by $P$ or $Q$ but not by $R$ is $(P \cap Q' \cap R') \cup (P' \cap Q \cap R') \cup (P \cap Q \cap R')$. Required probability $= P(P)P(Q')P(R') + P(P')P(Q)P(R') + P(P)P(Q)P(R')$. $= (\frac{3}{4} 	imes \frac{1}{2} 	imes \frac{3}{8}) + (\frac{1}{4} 	imes \frac{1}{2} 	imes \frac{3}{8}) + (\frac{3}{4} 	imes \frac{1}{2} 	imes \frac{3}{8})$. $= \frac{9}{64} + \frac{3}{64} + \frac{9}{64} = \frac{21}{64}$.

Three persons $P, Q$ and $R$ independently try to hit a target. If the probabilities of their hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$ respectively,then the probability that the target is hit by $P$ or $Q$ but not by $R$ is

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