Three students X, Y and Z appear for an exami | vedclass

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

Question

(A) Let $P(X), P(Y), P(Z)$ be the probabilities of students $X, Y, Z$ passing the exam respectively.Given:$P(X) = \frac{1}{5} \implies P(X') = 1 - \fr

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) Let $P(X), P(Y), P(Z)$ be the probabilities of students $X, Y, Z$ passing the exam respectively.Given:$P(X) = \frac{1}{5} \implies P(X') = 1 - \frac{1}{5} = \frac{4}{5}$$P(Y) = \frac{1}{4} \implies P(Y') = 1 - \frac{1}{4} = \frac{3}{4}$$P(Z') = \frac{2}{3} \implies P(Z) = 1 - \frac{2}{3} = \frac{1}{3}$We need the probability that at least two of them pass,which is the sum of probabilities of exactly two passing and all three passing.$P(	ext{at least 2 pass}) = P(X)P(Y)P(Z') + P(X)P(Y')P(Z) + P(X')P(Y)P(Z) + P(X)P(Y)P(Z)$$= (\frac{1}{5} 	imes \frac{1}{4} 	imes \frac{2}{3}) + (\frac{1}{5} 	imes \frac{3}{4} 	imes \frac{1}{3}) + (\frac{4}{5} 	imes \frac{1}{4} 	imes \frac{1}{3}) + (\frac{1}{5} 	imes \frac{1}{4} 	imes \frac{1}{3})$$= \frac{2}{60} + \frac{3}{60} + \frac{4}{60} + \frac{1}{60} = \frac{10}{60} = \frac{1}{6}$

Similar Questions

The probability of at least one of the events $A$ and $B$ occurring is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.2$, then $P(\overline{A}) + P(\overline{B})$ is

The probabilities of winning the race by two athletes $A$ and $B$ are $\frac{1}{5}$ and $\frac{1}{4}$ respectively. The probability that neither of them wins the race is:

One die of red colour,one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled,its colour and the number on its uppermost face is noted. Describe the sample space.

The probability of getting a total of $5$ or $6$ in a single throw of $2$ dice is

Vedclass Test Series

Exam Paper Generator

Online Exam Module