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: $1-p=1-1/2=1/2$

Let random variable $X=$ number of children having genotype AB: $X\sim B(n, p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given that $n=5, p=1/2$

 Find the probability that two of the children have genotype AB and three others have genotype BB .

$$P(X=2)=\binom{5}{2}({1 \over 2})^2(1-{1 \over 2})^{5-2}={5 \over 16}=0.3125$$