A man and his wife appear in an interview for two vacancies in the same post. The probability two 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?

Solution(By Examveda Team)

Let $${E_1}$$ = event that the husband is selected and $${E_2}$$ = event that the wife is selected.
Then, $$P({E_1}) = \frac{1}{7}{\text{ }}$$   and $${\text{ }}P({E_2}) = \frac{1}{5}$$
∴ $$P({\overline E _1}) = \left( {1 - \frac{1}{7}} \right)$$   $${\text{ = }}\frac{6}{7}$$
and
$$P({\overline E _2}) = $$   $$\left( {1 - \frac{1}{5}} \right) = \frac{4}{5}$$
∴ Required probability = P [( A and not B) or (B and not A)]
= P $$\left[ {\left( {{E_1} \cap {{\bar E}_2}} \right)\,{\text{or}}\,\left( {{E_2} \cap {{\bar E}_1}} \right)} \right]$$
= P $$\left( {{E_1} \cap {{\overline E }_2}} \right) + {\text{ P}}\left( {{E_2} \cap {{\overline E }_1}} \right)$$
= $${\text{P}}\left( {{E_1}} \right).{\text{ P}}\left( {{{\overline E }_2}} \right) + P\left( {{E_2}} \right).P\left( {{{\overline E }_1}} \right)$$
$$\eqalign{ & = \left( {\frac{1}{7} \times \frac{4}{5}} \right) + \left( {\frac{1}{5} \times \frac{6}{7}} \right) \cr & = \frac{10}{{35}} \cr & = \frac{2}{7} \cr} $$