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1. A man and his wife appear in an interview for two vacancies in the same post. The probability two 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. $\frac{8}{15}$

Solution:

Let

E_{1}

= event that the husband is selected and

E_{2}

= event that the wife is selected.
Then,

P (E_{1}) = \frac{1}{7}

and

P (E_{2}) = \frac{1}{5}

∴

P ({\bar{E}}_{1}) = (1 - \frac{1}{7})

= \frac{6}{7}

and

P ({\bar{E}}_{2}) =

(1 - \frac{1}{5}) = \frac{4}{5}

∴ Required probability = P [( A and not B) or (B and not A)]
= P

[(E_{1} \cap {\bar{E}}_{2}) or (E_{2} \cap {\bar{E}}_{1})]

= P

(E_{1} \cap {\bar{E}}_{2}) + P (E_{2} \cap {\bar{E}}_{1})

P (E_{1}) . P ({\bar{E}}_{2}) + P (E_{2}) . P ({\bar{E}}_{1})

\begin{aligned} = (\frac{1}{7} \times \frac{4}{5}) + (\frac{1}{5} \times \frac{6}{7}) \\ = \frac{10}{35} \\ = \frac{2}{7} \end{aligned}

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