binomial theorem with conditional probability There are four men in a room, 1 pair of brothers, and 2 unrelated men. The
probability that any man has blood-group X is
1
4
. The probability that if one
brother has blood-group X, the other brother also has X is
3
4
, otherwise the
blood-groups are independent. Find the probability that exactly 2 men in the
room have blood-group X.

$$\frac{3}{4}\cdot\left({3 \choose 2}\cdot\left(\frac{1}{4}\right)^2\cdot\frac{3}{4}\right)+\frac{1}{4}\cdot\left(\frac{3}{4}\cdot \left({2\choose 1}\frac{1}{4}\cdot\frac{3}{4}\right)+\frac{1}{4}\cdot\left(\frac{1}{4}\right)^2\right)=\frac{23}{128}$$
Chance of 1$^{\mathrm{st}}$ brother having X and not having X on the right and left, respectively.
