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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}$

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