Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have seven children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and four of the children are healthy. Round your answer to the nearest tenth.

The sickle cell anemia is the autosomal recessive condition means the trait will only be visible in the homozygous condition only. The parents are heterozygous ( Aa × Aa) and the progeny ( AA, Aa, Aa and aa) with the normal probability is 3/4 and the affected progeny probability is 1/4.

The probability of the normal and affected child can be calculated as follows by binomial theorem:

probability = n! / x! ( n! - x!) × pˣqⁿ⁻ˣ.

Here, n = total = 7, let x be the normal child and n! - x! is the affected child. p is the normal child probability = 3/4and q is affected child = 1/4.

Put the values in the given formula:

probability = [tex]\frac{7!}{4!3!} (3/4)^{4} (1/4)^{3}[/tex]

Probability = 2835 / 16384.

OR

probability = 17.3%.

Thus, the answer is 17.3%

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