If both husband and wife are known to be heterozygous for the autosomal recessive condition of albinism, what is the probability that among their four children, three will be normal and one will have albinism?

The albinism 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 albinism child can be calculated as follows:

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

Here, n = total = 4, 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{4!}{3!1!} (3/4)^{3} (1/4)^{1}[/tex]

probability = 27 / 64.

Thus, the answer is 27/64.