# A man and his wife appear for an interview for two posts. The probability of husband’s selection is $$\frac{1}{7}$$ and that of the wife’s selection is $$\frac{1}{5}$$ . Find the probability that only one of them will be selected.

Solution :
Let A be the event that husband is selected and B be the event that the wife is selected
then $P(A) =\large\frac{1}{7}$ and $P(B) =\large\frac{1}{5}$
$P(\bar {A}) =\bigg(1- \large\frac{1}{7} \bigg) =\large\frac{6}{7}$
$P(\bar {B}) =\bigg(1- \large\frac{1}{5} \bigg) =\large\frac{4}{5}$
Required probability is
$P(A) .P(\bar {B}) +P( \bar {A}) . P(B)$
$\qquad = \large\frac{1}{7} \times \large\frac{4}{5} +\large\frac{1}{5} \times \large\frac{6}{7}$
$\qquad= \large\frac{2}{7}$