find the probability of selecting exactly two women and at least two women when a six-person committee is selected from $7$ men and $4$ women? A committee of six members is formed from a group of $7$ men and $4$
women. What is the probability that the committee contains
a. exactly two women?
b. at least two women?
My attempt :  given $P(A) = 7/11$ and $P(B) = 4/11$
option a) probability that the commitee contains
exactly two women 
$$= \frac{P(AB)}{P(B)} = \frac{P(A)P(B)}{P(B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{\frac{4}{11}} = \frac{7}{11}$$
option b)the probability that the committee contains at least two women = 
$$\frac{P(AB)}{P(A \cup B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{1- \frac{7}{11} \cdot \frac{4}{11}} = \frac{28}{93}$$
If my answer is correct or not, I would be more thankful to those rectifying my mistakes......
 A: 
A committee of six members is formed from a group of $7$ men and $4$ women.  What is the probability that the committee contains exactly two women?

Since there are a total of $7 + 4 = 11$ people, the number of ways we can select a committee of six people is 
$$\binom{11}{6}$$
A committee of six that contains exactly two women must contain four of the seven men and two of the four women, so it can be selected in 
$$\binom{7}{4}\binom{4}{2}$$
ways.  Therefore, the probability that the committee contains exactly two women is 
$$\frac{\dbinom{7}{4}\dbinom{4}{2}}{\dbinom{11}{6}}$$

Under the same conditions as above, what is the probability that the committee contains at least two women?

If the committee contains at least two women, it must contain two, three, or four women.  Since there are six people on the committee, a committee that contains exactly $k$ women contains $6 - k$ of the seven men and $k$ of the four women.  Thus, there are 
$$\binom{7}{5}\binom{4}{2} + \binom{7}{4}\binom{4}{3} + \binom{7}{3}\binom{4}{4}$$
such committees, so the probability of selecting a committee with at least two women is 
$$\frac{\dbinom{7}{5}\dbinom{4}{2} + \dbinom{7}{4}\dbinom{4}{3} + \dbinom{7}{3}\dbinom{4}{4}}{\dbinom{11}{6}}$$
Alternatively, the number of committees that contain fewer than two women is 
$$\binom{7}{6}\binom{4}{0} + \binom{7}{5}\binom{4}{1}$$
so the probability that a committee contains fewer than two women is 
$$\frac{\dbinom{7}{6}\dbinom{4}{0} + \dbinom{7}{5}\dbinom{4}{1}}{\dbinom{11}{6}}$$
Thus, the probability that the committee contains at least two women is 
$$1 - \frac{\dbinom{7}{6}\dbinom{4}{0} + \dbinom{7}{5}\dbinom{4}{1}}{\dbinom{11}{6}}$$
An indication that you made an error is that the probability of selecting at least two women should be at least the probability that a committee contains exactly two women.
A: Isn't easier ton calculate the opposite probability for the second question 
b. at least two women?
Consider the $q$ probability that it contains $0$ woman or only $1$ woman then your probability is $p= 1-q$
@Taussig good answer but you did a little mistake it's not $2$ but $1$ in your formula.
$${{7}\choose{6}}{{4}\choose{0}} + {{7}\choose{5}}{{4}\choose{2}}$$
