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(B) Let $H$ be the event that the husband is selected and $W$ be the event that the wife is selected. Given $P(H) = \frac{1}{7}$ and $P(W) = \frac{1}{

Vedclass · Accepted Answer

$\frac{2}{7}$

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(B) Let $H$ be the event that the husband is selected and $W$ be the event that the wife is selected. Given $P(H) = \frac{1}{7}$ and $P(W) = \frac{1}{5}$. The probability that the husband is not selected is $P(H') = 1 - \frac{1}{7} = \frac{6}{7}$. The probability that the wife is not selected is $P(W') = 1 - \frac{1}{5} = \frac{4}{5}$. The probability that only one of them is selected is given by $P(	ext{only } H) + P(	ext{only } W)$. $P(	ext{only } H) = P(H) 	imes P(W') = \frac{1}{7} 	imes \frac{4}{5} = \frac{4}{35}$. $P(	ext{only } W) = P(W) 	imes P(H') = \frac{1}{5} 	imes \frac{6}{7} = \frac{6}{35}$. Therefore,the required probability is $\frac{4}{35} + \frac{6}{35} = \frac{10}{35} = \frac{2}{7}$.

$A$ man and his wife appear for an interview for two posts. The probability of the husband's selection is $\frac{1}{7}$ and that of the wife's selection is $\frac{1}{5}$. If they appear for the interview independently,then the probability that only one of them is selected is:

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