Conditional probability involving combinations A committee consists of 5 members.The 5 members are chosen from 9 men and 6 women.
Find the probability that 3 men and 2 women are chosen given that that there must be at least 2 women and 2 men in the committee.
Is the answer 9C3 × 6C2/(9C3×6C2 + 6C3 × 9C2)?
 A: Comment. Two exact computations and a simulation in R statistical software:
Sometimes one gains confidence in an answer by looking at the problem in several different ways.  
(1) Numerical computation of your result:
choose(9,3)*choose(6,2)/(choose(9,3)*choose(6,2)+choose(9,2)*choose(6,3))
## 0.6363636

(2) Let $W$ be the hypergeometric random variable that counts the number
of women selected in this experiment. You want 
$$P(W = 2 | W =2\; or\; 3) = P(W=2)/[P(W = 2) + P(W=3)].$$
dhyper(2, 6, 9, 5)/sum(dhyper(2:3, 6, 9, 5))
## 0.6363636

(3) Simulate the experiment a million times and count the proportion of instances of $\{W=2\}$ among instances of $\{W=2\} \cup \{W = 3\}.$
Approximation with about three place accuracy. (In R, | signifies $\cup$,
and the mean of a logical vector of TRUEs and FALSEs is its
proportion of TRUEs.)
m = 10^6;  w = numeric(m)
pop = c(rep(1, 6), rep(0,9))  # six women (1's) and nine men
for (i in 1:m) {  w[i] = sum(sample(pop, 5))  }
mean(w[w==2|w==3]==2)
## 0.6358527

