Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
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 ?
(a)$\frac{4}{5}$
(b)$\frac{2}{7}$
(c)$\frac{4}{7}$
(d)$\frac{8}{15}$
(e)None of these
Answer
Answer (as printed): B
Explanation
Let $E_{1}=$ Event that the husband is selected and $E_{2}=$ Event that the wife is selected. Then, $$\begin{aligned} & P\left(E_{1}\right)=\frac{1}{7} \text { and } P\left(E_{2}\right)=\frac{1}{5} . \\ \therefore & P\left(\bar{E}_{1}\right)=\left(1-\frac{1}{7}\right)=\frac{6}{7} \text { and } P\left(\bar{E}_{2}\right)=\left(1-\frac{1}{5}\right)=\frac{4}{5} . \\ \therefore & \text { Required probability }=P[(A \text { and not } B) \text { or }(B \text { and not } A)] \\ = & P\left[\left(E_{1} \cap \bar{E}_{2}\right) \text { or }\left(E_{2} \cap \bar{E}_{1}\right)\right] \\ = & P\left[\left(E_{1} \cap \bar{E}_{2}\right)+P\left(E_{2} \cap \bar{E}_{1}\right)\right] \\ = & P\left(E_{1}\right) \cdot P\left(\bar{E}_{2}\right)+P\left(E_{2}\right) \cdot P\left(\bar{E}_{1}\right)=\left(\frac{1}{7} \times \frac{4}{5}\right)+\left(\frac{1}{5} \times \frac{6}{7}\right)=\frac{10}{35}=\frac{2}{7} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.