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(B) Given probabilities of hitting the target are $P(P) = \frac{3}{4}$,$P(Q) = \frac{1}{2}$,and $P(R) = \frac{5}{8}$.Since the events are independent,

Vedclass · Accepted Answer

$\frac{21}{64}$

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(B) Given probabilities of hitting the target are $P(P) = \frac{3}{4}$,$P(Q) = \frac{1}{2}$,and $P(R) = \frac{5}{8}$.Since the events are independent,the probabilities of not hitting the target are $P(P') = 1 - \frac{3}{4} = \frac{1}{4}$,$P(Q') = 1 - \frac{1}{2} = \frac{1}{2}$,and $P(R') = 1 - \frac{5}{8} = \frac{3}{8}$.We need the probability that the target is hit by $P$ or $Q$ but not by $R$. This corresponds to the event $(P \cap Q' \cap R') \cup (P' \cap Q \cap R') \cup (P \cap Q \cap R')$.Since these are mutually exclusive cases,the probability is:$P = P(P)P(Q')P(R') + P(P')P(Q)P(R') + P(P)P(Q)P(R')$$P = \left(\frac{3}{4} 	imes \frac{1}{2} 	imes \frac{3}{8}ight) + \left(\frac{1}{4} 	imes \frac{1}{2} 	imes \frac{3}{8}ight) + \left(\frac{3}{4} 	imes \frac{1}{2} 	imes \frac{3}{8}ight)$$P = \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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