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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 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 events are independent,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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