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……
Best Answer
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 $$\frac{\dbinom{7}{4}\dbinom{4}{2}}{\dbinom{11}{6}}$$
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.