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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$\frac{21}{64}$

Vedclass · Answer

(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

Similar Questions

In a game of throwing a die,what is the probability of getting a $1$ on an even-numbered throw?

Two persons $A$ and $B$ take turns in throwing a pair of dice. The first person to throw a sum of $9$ with both dice will win the game. If $A$ throws first,then the probability that $B$ wins the game is:

$A, B, C$ are mutually exclusive events such that $P(A) = \frac{3x+1}{3}$, $P(B) = \frac{1-x}{4}$, and $P(C) = \frac{1-2x}{2}$. Then the set of possible values of $x$ is:

$A$ dice marked with digits $\{1, 2, 2, 3, 3, 3\}$ is thrown three times. The probability that the sum of the numbers on the faces is $6$ is equal to:

If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P(F) < 1,$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module