woman with achondroplasia a dominant form of dwarfism and a

woman with achondroplasia (a dominant form of dwarfism) and a phenotypically unaffected man have seven children, all of whom have achondroplasia. What is the probability of producing such a family if this woman is a heterozygote? What is the probability that the woman is a heterozygote if her eighth child does not have this disorder?

Solution

Let\'s begin with a very easy case: question about gender probabilities in families of eight children.

The probability that all eight children in a family will be the identical gender
P(all female)= 1/2 x 1/2 x 1/2 x 1/2 x1/2 x 1/2 x 1/2 x 1/2= 1/8
P(all male ) = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/8
P(all one gender) = P(all female) + P(all male) = 1/8 + 1/8 = 1/4

The probability that a eight -child family is 6 girls and 2 boys
Each feasible birth sequence has P=1/8. That is, P(G,G,B)=P(G,B,G)=P(B,G,G)=1/8.
So, P(2G,1B)= 3/8 and P(1G,2B)= 3/8.

This permit us to write the total gender probability dispensation for families of eight children as accompanyed:
Eight girls will be 1/8
four girls will be 3/8 and two boys
Two girls will be 3/8 and two boys
1/8 will be Eight boys
Connect it all up, we have 1/8 + 3/8 + 3/8 + 1/8 = 1 (100%)