Question

There are five children in a family of parents AB× BB. The children of such parents must
have genotype AB or genotype BB . Find the probability that two of the children have
genotype AB and three others have genotype BB .

Answer 1

Let probability of a child having genotype AB =p=1/2.

Then probability of a child having genotype BB=1-p=1/2.

Let random variable X = number of children having genotype AB in 5 children i.e. in 5 trials 

$$X\sim B(n, p), n=5, p=1/2$$

$$P(X=2)=\begin{pmatrix} 5 \\ 2 \end{pmatrix}({1 \over 2})^2(1-{1 \over 2})^{5-2}={5! \over {2!3!}}\cdotp {1 \over 32}={5 \over 16}$$