binomial theorem with conditional probability

Question

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.

Answer 1

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