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

Vedclass · Accepted Answer

$\frac{2}{7}$

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(A) Let $A$ be the event that the husband is selected and $B$ be the event that the wife is selected.Given: $P(A) = \frac{1}{7}$ and $P(B) = \frac{1}{5}$.Then,the probability of not being selected is:$P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{7} = \frac{6}{7}$$P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$The probability that only $1$ of them is selected is given by the case where the husband is selected and the wife is not,$OR$ the wife is selected and the husband is not:$P(	ext{only one}) = P(A) \cdot P(\bar{B}) + P(B) \cdot P(\bar{A})$$= \left(\frac{1}{7} 	imes \frac{4}{5}ight) + \left(\frac{1}{5} 	imes \frac{6}{7}ight)$$= \frac{4}{35} + \frac{6}{35} = \frac{10}{35} = \frac{2}{7}$

$A$ husband and wife appear in an interview for $2$ vacancies in the same post. The probability of the husband's selection is $\frac{1}{7}$ and that of the wife's is $\frac{1}{5}$. What is the probability that only $1$ of them will be selected?

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