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 wife’s selection is $\frac{1}{5}$. What is the probability that only one of them will be selected.

$\frac{2}{7}$

The probability of husband is not selected $=1-\frac{1}{7}=\frac{6}{7}$

The probability that wife is not selected $=1-\frac{1}{5}=\frac{4}{5}$

The probability that only husband is selected $=\frac{1}{7}× \frac{4}{5}=\frac{4}{35}$

The probability that only wife is selected $=\frac{1}{5}× \frac{6}{7}=\frac{6}{35}$

Hence, required probability $=\frac{6}{35}+\frac{4}{35}=\frac{10}{35}=\frac{2}{7}.$