# 4 member committees with more women than men

From a group of 10 men and 5 women, 4 member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men is:

1. $\frac{2}{23}$
2. $\frac{1}{11}$
3. $\frac{21}{220}$
4. $\frac{3}{11}$

Since each committee must have at least one woman, 4 women have already been placed into separate committees. There is one woman yet to be placed. For a committee to have more women than men, there can be 2 women and 0 men or 1 man. For a committee having only one woman, there must not be any more men in the committee.

This is what I could reason so far. But I don't know how to proceed.

Edit: I might have misinterpreted four-member committees (having 4 members) as 4 committees.

• "the probability for these committees to have more women than men" - That's awfully vague. For at least one of these committees to contain more women than men? For all of them to contain more women than men? For one chosen at random to have this property? All these interpretations would give different results. – TMM Apr 10 '17 at 18:19
• @TMM That's exactly my point. I was totally confused during the exam. Hence I did not attempt it ultimately. (There is negative marking.) – Hungry Blue Dev Apr 10 '17 at 18:30
• Are you quoting the exact question then? If so, that's pretty bad on the examiners side. – TMM Apr 10 '17 at 20:10
• Yes. It is the exact question. The exam took place on Sunday 9th April (and 8th as well): The online version of JEE Mains 2017. – Hungry Blue Dev Apr 10 '17 at 20:12

The probability that there is at least one woman in the committee is 1 minus the probability that no woman is in the committee

$$P(W\geq 1)=1- \frac{{10 \choose 4}\cdot {5 \choose 0}}{15 \choose 4}$$

Then it is asked for $$P(W\geq 3|W\geq 1)=\frac{P(W\geq 3\cap W\geq 1)}{P(W \geq 1)}=\frac{P(W\geq 3)}{P(W\geq 1)}$$

And $$P(W \geq 3)=\sum_{x=3}^4 \frac{{10 \choose 4-x}\cdot {5 \choose x}}{{15 \choose 4}}$$

• In the last line... shouldn't it be $\sum_{x=3}^{4}$ instead of $\sum_{x=3}^{5}$? – Hungry Blue Dev Apr 10 '17 at 17:55
• Yes you´re right. – callculus Apr 10 '17 at 18:01

each group should have at least one women so total no of ways $=\binom{15}{4}-\binom{10}{4}$

$$P(W \ge 3) =\frac{\binom{5}4 \binom{10}0+\binom{5}3\binom{10}1}{\binom{15}4-\binom{10}4}$$

If you consider each committee to contain 4 members and assume that there are 4 committees,

P(4 member committees which contain at least 1 woman) = P(3M,1W)+P(2M,2W)+P(1M,2W)+P(0M,4W) =

{10C3.5C1 + 10C2.5C2 + 10C1.5C3 + 10C0.5C4}/ 15C4 = 1155/1365

P(committees to have more women than men) =

{ P(1M,3W) + P(0M,4W) }/ {P(3M,1W)+P(2M,2W)+P(1M,2W)+P(0M,4W)}

=(105/1365)/(1155/1365) = 1/11, which is the required probability.