Question

A husband and wife appeared for the interview of the some posts. If the probability of the husband selected is $\frac{1}{7}$ and that of the wife is $\frac{1}{5}$. The probability of one of them selected is:

• A. $\frac{24}{35}$
• B. $\frac{11}{35}$
• C. $\frac{10}{35}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$\frac{10}{35}$

Let A= event that the husband is selected.

and B = event that the wife is selected.

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

Therefore, $P\left(\overline{A}\right)=\frac{6}{7},P\left(\overline{B}\right)=\frac{4}{5}$

Therefore, the required probability ofnone of them selected = P(not A and B) or P(A and not B)

$⟹P\left(\overline{A}\right).P\left(B\right)+P\left(A\right).P\left(\overline{B}\right)=(\frac{6}{7}\times \frac{1}{5})+(\frac{1}{7}\times \frac{4}{5})=\frac{10}{35}$