A man and his wife appear for an interview for two posts. The probability of the husband's selection is…

A man and his wife appear for an interview for two posts. The probability of the husband's selection is $\frac{1}{7}$. and that of the wife's selection is $\frac{1}{5}$. If they appear for the interview independently, then the probability that only one of them is selected, is
  1. $\frac{1}{7}$
  2. $\frac{2}{7}$
  3. $\frac{6}{7}$
  4. $\frac{4}{7}$

Solution

The probability of husband is not selected $=1-\frac{1}{7}=\frac{6}{7}$
The probability that wife is not selected $=1-\frac{1}{5}=\frac{4}{5}$
The probability that only husband selected $=\frac{1}{7} \times \frac{4}{5}=\frac{4}{35}$
The probability that only wife selected $=\frac{1}{5} \times \frac{6}{7}=\frac{6}{35}$ $\begin{aligned} \text { Hence, required probability } & =\frac{6}{35}+\frac{4}{35}=\frac{10}{35} \\ & =\frac{2}{7} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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