Question

Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid, "sickle" shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 3 children, what is the probability that two will have the disease?

Answer 1

Answer:

The probability that two will have the disease is 0.1406.

Explanation:

Let X = number of children having the disease.

The probability of a children having the disease is, p = 0.25.

The number of children a couple has, n = 3.

A children having the disease is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 3 and p = 0.25.

The probability mass function of the binomial random variable X is:

$P(X=x)={3\choose x}0.25^{x}(1-0.25)^{3-x};\ x=0,1,2,3...$

Compute the probability that of the 3 children, 2 will have the disease is:

$P(X=2)={3\choose 2}0.25^{2}(1-0.25)^{3-2}\\=3\times 0.0625\times 0.75\\=0.140625\\\approx0.1406$

Thus, the probability that two will have the disease is 0.1406.