The probability of hitting a target by three | vedclass

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

Question

(A) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.Then $P(\overline{A}) = \frac{1}{2}, P(\overline{B}) = \frac{2}{3}, P(\overline{C

Vedclass · Accepted Answer

$\frac{13}{24}$

Vedclass · Answer

(A) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.Then $P(\overline{A}) = \frac{1}{2}, P(\overline{B}) = \frac{2}{3}, P(\overline{C}) = \frac{3}{4}$.$\lambda$ is the probability that exactly two hit the target:$\lambda = P(A)P(B)P(\overline{C}) + P(A)P(\overline{B})P(C) + P(\overline{A})P(B)P(C)$$\lambda = (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{3}{4}) + (\frac{1}{2} 	imes \frac{2}{3} 	imes \frac{1}{4}) + (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{4})$$\lambda = \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$.$\mu$ is the probability that at least two hit the target:$\mu = \lambda + P(A)P(B)P(C)$$\mu = \frac{6}{24} + (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{4}) = \frac{6}{24} + \frac{1}{24} = \frac{7}{24}$.$\lambda + \mu = \frac{6}{24} + \frac{7}{24} = \frac{13}{24}$.

The probability of hitting a target by three marksmen is $\frac{1}{2}, \frac{1}{3},$ and $\frac{1}{4}$ respectively. If the probability that exactly two of them will hit the target is $\lambda$ and that at least two of them hit the target is $\mu,$ then $\lambda + \mu$ is equal to :-

Similar Questions

If $A$ and $B$ are events of a random experiment such that $P(A \cup B) = \frac{4}{5}$,$P(\bar{A} \cup \bar{B}) = \frac{7}{10}$,and $P(B) = \frac{2}{5}$,then $P(A)$ equals

$A$ problem in mathematics is given to $4$ students whose chances of solving it individually are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4},$ and $\frac{1}{5}.$ The probability that the problem will be solved by at least one student is

The probability of obtaining an even prime number on each die,when a pair of dice is rolled is

The probability of happening an event $A$ is $0.5$ and that of $B$ is $0.3$. If $A$ and $B$ are mutually exclusive events,then the probability of happening neither $A$ nor $B$ is

$A$ die is rolled. Let $E$ be the event "die shows $4$" and $F$ be the event "die shows an even number". Are $E$ and $F$ mutually exclusive?

Vedclass Test Series

Exam Paper Generator

Online Exam Module