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Question

(a) Let ${E_1}$ be the event that man will be selected and ${E_2}$ the event that woman will be selected. Then

$P({E_1}) = \frac{1}{4}$

so $P({\

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(a) Let ${E_1}$ be the event that man will be selected and ${E_2}$ the event that woman will be selected. Then

$P({E_1}) = \frac{1}{4}$

so $P({\bar E_1}) = 1 - \frac{1}{4} = \frac{3}{4}$ and $P({E_2}) = \frac{1}{3}$

So $P({\bar E_2}) = \frac{2}{3}$

Clearly ${E_1}$ and ${E_2}$ are independent events.

So, $P({\bar E_1} \cap {\bar E_2}) = P({\bar E_1}) 	imes P({\bar E_2}) = \frac{3}{4} 	imes \frac{2}{3} = \frac{1}{2}$.

A man and a woman appear in an interview for two vacancies in the same post. The probability of man's selection is $1/4$ and that of the woman's selection is $1/3$. What is the probability that none of them will be selected

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