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Question

(A) Let $E_1$ be the event that the man is selected and $E_2$ be the event that the woman is selected.Given,$P(E_1) = 1/4$ and $P(E_2) = 1/3$.The prob

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$1/2$

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(A) Let $E_1$ be the event that the man is selected and $E_2$ be the event that the woman is selected.Given,$P(E_1) = 1/4$ and $P(E_2) = 1/3$.The probability that the man is not selected is $P(\bar{E}_1) = 1 - P(E_1) = 1 - 1/4 = 3/4$.The probability that the woman is not selected is $P(\bar{E}_2) = 1 - P(E_2) = 1 - 1/3 = 2/3$.Since the selection of the man and the woman are independent events,the probability that none of them will be selected is $P(\bar{E}_1 \cap \bar{E}_2) = P(\bar{E}_1) 	imes P(\bar{E}_2)$.Therefore,$P(\bar{E}_1 \cap \bar{E}_2) = 3/4 	imes 2/3 = 6/12 = 1/2$.

$A$ man and a woman appear in an interview for two vacancies in the same post. The probability of the man's selection is $1/4$ and that of the woman's selection is $1/3$. What is the probability that none of them will be selected?

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