Question

Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is $\frac{1}{7}$ and that of John's selection is $\frac{1}{5}$. 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

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