Two candidates, X and Y, appeared in an interview for two vacancies for the same post. The probabilities of their selection are 1/4 and 1/7, respectively. What is the probability that one of them will be selected?

9/28

The correct answer is Option (4) → 9/28

Let the probabilities of selection be:

• $P(X) = \frac{1}{4}$​
• $P(Y) = \frac{1}{7}$​

Since there are two vacancies, both candidates can be selected.
The question asks for the probability that exactly one of them is selected.

$P(\text{exactly one}) = P(X \text{ selected, } Y \text{ not}) + P(Y \text{ selected, } X \text{ not})$

$= \frac{1}{4} \times \frac{6}{7} + \frac{1}{7} \times \frac{3}{4}$

$= \frac{6}{28} + \frac{3}{28} = \frac{9}{28}$