Question

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

Answer 1

Hint: In this question, we need to determine the probability that a particular man and a particular woman would both be on the committee such that the committee is made by randomly selecting candidates from a group of 9 men and 11 women. For this, we will use the concept of combinations to select 3 men and women from the group and then, for the particular man and the woman.

Complete step-by-step answer:
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 \[ = {}^9{C_3} \times {}^{11}{C_3} = \dfrac{{9!}}{{3! \times 6!}} \times \dfrac{{11!}}{{3! \times 8!}} = 84 \times 165 = 13860\]
Also 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
\[ = {}^{20}{C_6} = \dfrac{{20!}}{{14! \times 6!}} = 38760\]
Therefore the probability of selecting 3 men and 3 women will be
\[ = \dfrac{{13860}}{{38760}} = \dfrac{{6930}}{{19380}} = \dfrac{{2310}}{{6460}} = \dfrac{{231}}{{646}}\]

Note: \[{}^n{C_r}\], is the mathematical representation of the combination which is a method of selection of some items or all of the items from a set without considering the sequence of selection whereas in the case of permutation which is the method of arrangements of items of a set the sequence is considered represented as \[{}^n{P_r}\].
\[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] and \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]