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A man and a woman appear in an interview for two vacancies in the same post. The probability of man’s selection is \[\dfrac{1}{4}\] and probability of woman’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

Last updated date: 25th Feb 2024
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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.

Complete step-by-step answer:
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 \[P({E_1}) = \dfrac{1}{4}\].
∴ Probability of man not getting selected is \[P({\bar E_1}) = 1 - \dfrac{1}{4}\]
Similarly, probability of woman getting selected is \[P({E_2}) = \dfrac{1}{3}\]
∴ 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
\[P({\bar E_1} \cap {\bar E_2}) = P({\bar E_1}) \times P({\bar E_2})\]
=\[\dfrac{3}{4} \times \dfrac{2}{3}\]
∴ The correct option is ‘a’.

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 \[{E_2}\] \[P({E_1} \cap {E_2}) = P({E_1}) \times P({E_2})\]