Question

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

Answer 1

Let $A=$ be the event that the husband is selected.

and $A=$ be the event that the wife is selected.

Then $P\left(A\right)=\frac{1}{7}$ and $P\left(B\right)=\frac{1}{5}$

$\therefore P\left(\overline{A}\right)=1-\frac{1}{7}=\frac{7-1}{7}=\frac{6}{7}$

and $P\left(\overline{B}\right)=1-\frac{1}{5}=\frac{5-1}{5}=\frac{4}{5}$

$\therefore$ Required probability$=P\left[\left(AandnotB\right)or\left(BandnotA\right)\right]$

$=P[\left(Aand\overline{B}\right)+\left(Band\overline{A}\right)]$

$=P\left(A\right)P\left(\overline{B}\right)+P\left(B\right)P\left(\overline{A}\right)$

$=\frac{1}{7}\times \frac{4}{5}+\frac{1}{5}\times \frac{6}{7}$

$=\frac{4+6}{35}=\frac{10}{35}=\frac{2}{7}$