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.
 A: 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 $$P(W \geq 3)=\sum_{x=3}^4 \frac{{10 \choose 4-x}\cdot {5 \choose x}}{{15 \choose 4}}$$
A: each group should have at least one women  so total no of ways $=\binom{15}{4}-\binom{10}{4}$
$$P(W \ge 3) =\frac{\binom{5}4 \binom{10}0+\binom{5}3\binom{10}1}{\binom{15}4-\binom{10}4}$$
A: 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.
