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


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