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Question: A man and a woman appear in an interview for two vacancies in the same post. The probability of man'...

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

A

12\frac { 1 } { 2 }

B

112\frac { 1 } { 12 }

C

14\frac { 1 } { 4 }

D

None of these

Answer

12\frac { 1 } { 2 }

Explanation

Solution

Let E1E _ { 1 } be the event that man will be selected and E2E _ { 2 } the event that woman will be selected. Then

P(E1)=14P \left( E _ { 1 } \right) = \frac { 1 } { 4 } so P(Eˉ1)=114=34P \left( \bar { E } _ { 1 } \right) = 1 - \frac { 1 } { 4 } = \frac { 3 } { 4 } and P(E2)=13P \left( E _ { 2 } \right) = \frac { 1 } { 3 }

So P(Eˉ2)=23P \left( \bar { E } _ { 2 } \right) = \frac { 2 } { 3 }

Clearly E1E _ { 1 } and E2E _ { 2 } are independent events.

So, P(Eˉ1Eˉ2)=P(Eˉ1)×P(Eˉ2)=34×23=12P \left( \bar { E } _ { 1 } \cap \bar { E } _ { 2 } \right) = P \left( \bar { E } _ { 1 } \right) \times P \left( \bar { E } _ { 2 } \right) = \frac { 3 } { 4 } \times \frac { 2 } { 3 } = \frac { 1 } { 2 } .