Question

Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is 1/3 and that of Monika's selection is 1/5. Find the probability that
(i) both of them will be selected
(ii) none of them will be selected
(iii) at least one of them will be selected
(iv) only one of them will be selected.

Answer 1

$$\begin{gathered} P\left(\text{Kamal gets selected}\right)=P\left(A\right)=\frac{1}{3} \\ P\left(\text{Monica gets selected}\right)=P\left(B\right)=\frac{1}{5} \\ \left(\mathrm{i}\right)P\left(\text{both get selected}\right)=P\left(A\right)\times P\left(B\right) \\ =\frac{1}{3}\times \frac{1}{5} \\ =\frac{1}{15} \\ \left(\mathrm{ii}\right)P\left(\text{none of them get selected}\right)=P\left(\overline{A}\right)\times P\left(\overline{B}\right) \\ =\left[1-P\left(A\right)\right]\left[1-P\left(B\right)\right] \\ =\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right) \\ =\frac{2}{3}\times \frac{4}{5} \\ =\frac{8}{15} \\ \left(\mathrm{iii}\right)P\left(\text{atleast one of them gets selected}\right)=P\left(A\cup B\right) \\ =P\left(A\right)+P\left(B\right)-P\left(A\cap B\right) \\ =P\left(A\right)+P\left(B\right)-P\left(A\right)\times P\left(B\right) \\ =\frac{1}{3}+\frac{1}{5}-\frac{1}{3}\times \frac{1}{5} \\ =\frac{1}{3}+\frac{1}{5}-\frac{1}{15} \\ =\frac{7}{15} \\ \left(\mathrm{iv}\right)P\left(\text{one of them gets selected}\right)=P\left(\overline{A}\right)P\left(B\right)+P\left(\overline{B}\right)P\left(A\right) \\ =P\left(B\right)\left[1-P\left(A\right)\right]+P\left(A\right)\left[1-P\left(B\right)\right] \\ =\frac{1}{5}\left(1-\frac{1}{3}\right)+\frac{1}{3}\left(1-\frac{1}{5}\right) \\ =\frac{2}{15}+\frac{4}{15} \\ =\frac{6}{15}=\frac{2}{5} \end{gathered}$$