Question

# A man and a woman appear in an interview for two vacancies in the same post. The probability of manâ€™s selection is $\dfrac{1}{4}$ and probability of womanâ€™s selection is $\dfrac{1}{3}$. What is the probability that none of them will be selected?A) $\dfrac{1}{2}$B) $\dfrac{1}{{12}}$C) $\dfrac{1}{4}$D) None of these

Hint: First look at the condition asked, then find the probability of the man getting selected, then from that find the probability of not getting a man selected. Similarly, find the probability of getting a woman selected and the probability of women not getting selected. And then proceed for the final step.

Let ${E_1}$ be the event that man will be selected and ${E_2}$ be the event that woman will be selected.
Thus, the probability of a man getting selected is $P({E_1}) = \dfrac{1}{4}$.
âˆ´ Probability of man not getting selected is $P({\bar E_1}) = 1 - \dfrac{1}{4}$
Similarly, probability of woman getting selected is $P({E_2}) = \dfrac{1}{3}$
âˆ´ Probability of woman not getting selected is $P({\bar E_2}) = 1 - \dfrac{1}{3}$
Now the probability of none getting selected is given by
$P({\bar E_1} \cap {\bar E_2}) = P({\bar E_1}) \times P({\bar E_2})$
=$\dfrac{3}{4} \times \dfrac{2}{3}$
=$\dfrac{1}{2}$
âˆ´ The correct option is â€˜aâ€™.

Note: In this question the three events are of independent nature, the general formula for their probability is given by the formula below, say these two events are ${E_1}$ and ${E_2}$ $P({E_1} \cap {E_2}) = P({E_1}) \times P({E_2})$