A man and woman are both heterozygous for a dominant disorde

A man and woman are both heterozygous for a dominant disorder. What is the probability that if they have 5 children, 3 will have the disorder?

Answer: 31/64

Explanation:

If both parents were heterozygotes, the probability of disorder is 3/4 and normal possibility is 1/4

Aa x Aa

AA (1/2) & Aa (1/2) = A_ (3/4, disorder)

aa (1/4 non disorder)

P(3disorder, 2normal) = P(3/4 * 3/4 * 3/4) + (1/4 + 1/4)

= 27/64 + 1/16 = 31/64