Question

A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is 1/7 and that of wife's selection is 1/5. What is the probability that
(i) both of them will be selected?
(ii) only one of them will be selected?
(iii) none of them will be selected?

Answer 1

$$\begin{gathered} P\left(\text{husband will be selected}\right)=P\left(A\right)=\frac{1}{7} \\ P\left(\text{wife will be selected}\right)=P\left(B\right)=\frac{1}{5} \\ \left(\mathrm{i}\right)P\left(\text{both will be selected}\right)=P\left(A\cap B\right) \\ =P\left(A\right)\times P\left(B\right) \\ =\frac{1}{7}\times \frac{1}{5} \\ =\frac{1}{35} \\ \left(\mathrm{ii}\right)P\left(\text{only one of them will be selected}\right)=P\left(A\right)P\left(\overline{B}\right)+P\left(\overline{A}\right)P\left(B\right) \\ =\frac{1}{7}\left(1-\frac{1}{5}\right)+\frac{1}{5}\left(1-\frac{1}{7}\right) \\ =\frac{4}{35}+\frac{6}{35} \\ =\frac{10}{35} \\ =\frac{2}{7} \\ \left(\mathrm{iii}\right)P\left(\text{none of them will be selected}\right)=P\left(\overline{A}\cap \overline{B}\right) \\ =P\left(\overline{A}\right)\times P\left(\overline{B}\right) \\ =\left(1-\frac{1}{7}\right)\left(1-\frac{1}{5}\right) \\ =\frac{24}{35} \end{gathered}$$