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

### Concept & Logic When two independent events occur, the probability that exactly one happens is the sum of two scenarios: (Event A happens AND Event B does not) OR (Event A does not happen AND Event B happens). $$ P(\text{only one}) = P(A) \times P(\bar{B}) + P(\bar{A}) \times P(B) $$ ### Step-by-Step Solution * Let $H$ = husband is selected, and $W$ = wife is selected. * $P(H) = \frac{1}{7}$ * $P(\bar{H}) = 1 - P(H) = 1 - \frac{1}{7} = \frac{6}{7}$ (Probability husband is not selected) * $P(W) = \frac{1}{5}$ * $P(\bar{W}) = 1 - P(W) = 1 - \frac{1}{5} = \frac{4}{5}$ (Probability wife is not selected) * The required probability is $P(\text{only one selected}) = P(\text{Husband selected and Wife not}) + P(\text{Husband not selected and Wife selected})$ * $= [P(H) \times P(\bar{W})] + [P(\bar{H}) \times P(W)]$ * $= \left(\frac{1}{7} \times \frac{4}{5}\right) + \left(\frac{6}{7} \times \frac{1}{5}\right)$ * $= \frac{4}{35} + \frac{6}{35} = \frac{10}{35}$ * Simplifying the fraction: $\frac{10}{35} = \frac{2}{7}$ ### Exam Strategy & Shortcut Write down the direct and complementary probabilities immediately: $(H, \bar{H}) = (1/7, 6/7)$ and $(W, \bar{W}) = (1/5, 4/5)$. Cross-multiply the success of one with the failure of the other and sum them up: $(1 \times 4 + 6 \times 1) / 35 = 10 / 35 = 2/7$. ### Common Pitfall A common pitfall is calculating the probability that *at least* one is selected ($1 - P(\text{none})$) instead of *exactly* one. ### Final Answer Therefore, the correct answer is **$\frac{2}{7}$**.