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If a committee of 3 men and 3 women is randomly selected from a group of 9 men and 11 women, then the probability that a particular man and a particular women would both be on the committee, is

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\frac{213}{646}

\frac{221}{646}

\frac{231}{646}

\frac{237}{646}

answer is C.

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Detailed Solution

Number of men in a group =9
Number of women in a group =11
Now the number of ways of selecting 3 men and 3 women from a group of 9 men and 11 women will be

C_{3}^{9} \times C_{3}^{11} = 13860

The number of ways of selecting 6 persons (3 men + 3 women) from the group of 9 men and 11 women will be equal to

C_{6}^{20} = 38760

Probability =

\frac{13860}{38760} = \frac{231}{646}

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