Question

A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband‘s selection is \(\frac{1}{7}\) and that of wife's selection is \(\frac{1}{5}\) . What is the probability that 

(i) Both of them will be selected 

(ii) Only one of them will be selected 

(iii) None of them will be selected 

(iv) At least one of them will be selected.

Answer 1

Let A : Event of husband being selected 

B : Event of wife being selected

Then P(A) = \(\frac{1}{7},\)   P(B) = \(\frac{1}{5}\)

P(\(\bar{A}\)) = 1 - \(\frac{1}{7}\) = \(\frac{6}{7}\), P(\(\bar{B}\)) = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)

(i) P(Both are selected) = P(A) × P(B) = \(\frac{1}{7}\) x \(\frac{1}{5}\) = \(\frac{1}{35}\)

(ii) P(only one is selected) = P(A selected) × P(B not selected) + P(A not selected) × P(B selected)

= P(A) x P(\(\bar{B}\)) + P(\(\bar{A}\)) x P(B) = \(\frac{1}{7}\) x \(\frac{4}{5}\) + \(\frac{6}{7}\) x \(\frac{1}{5}\) = \(\frac{4}{35}\) + \(\frac{6}{35}\) = \(\frac{10}{35}\) = \(\frac{2}{7}\)

(iii) P(none selected) = P(\(\bar{A}\)) x P(\(\bar{B}\)) = \(\frac{6}{7}\) x \(\frac{4}{5}\) = \(\frac{24}{35}\)

(iv) P(at least one selected) = 1 – P(none selected) = 1 - \(\frac{24}{35}\) = \(\frac{11}{35}\)