From a group consisting of three women and five men, a committee of four people has to be chosen. The assumption is that each person has the same probability to be chosen and can be chosen only once.
Determine the probabilities that the committee consists of
(1) at least two women
(2) at most one man.
I would think that it makes sense to assume four random variables $X_i\in \{0,1\}, i=1,2,3,4$, where the event $X_i=0$ is meant to mean "Person $i$ is a man" and $X_i=1$ is supposed to mean "Person $i$ is a woman". For each of the four RV I suppose it is Bernoulli distributed with $p=\frac{3}{8}$ and $1-p=\frac{5}{8}$.
Moreover, I let $S_4=X_1+X_2+X_3+X_4$.
Then my solutions are:
(1) $P(S_4\geqslant 2)=1-P(S_4<2)=1-P(S_4=0)-P(S_4=1)=1-\binom{4}{0}p^0(1-p)^4-\binom{4}{1}p(1-p)^3\approx 0.49$
(2) $P(S_4\geqslant 3)\approx 0.49-\binom{4}{2}p^2(1-p)^2\approx 0.16$
Since it is long time ago that I solved this kind of exercises, I would like to know whether my solution makes sense.
In a problem like formung a committee, you need to realize that it is selection without replacement so you can't use the binomial distribution formula, you have to use the hypergeometric formula, eg for part $1$,
$$\frac{\binom32\binom52 + \binom33\binom51}{\binom84}$$
you should now be able to work out the second problem