Polydactyly is expressed when an individual has extra fingers and/or toes. Assume that a man with 6 fingers on each hand and 6 toes on each foot marries a woman with a normal number of digits. Having extra digits is caused by a dominant allele. The couple has a son with normal hands and feet, but the couple's second child has extra digits.
a)What is the probability that their next child will have polydactyly? b)What is the probability that their next 3 children will not have polydactyly? c)What is the probability that , of their next 3 offspring, any 2 will have polydactyly and the other one will not?

Polydactyly is an autosomal dominant trait i.e. the progeny will acquire the disease even if he/she has one dominant allele. From the conditions mentioned in the question, it is clear that father is genotypically heterozygous because if father was homozygous dominant for the trait then not even a single progeny would have born normal.

Father Mother

Pp x pp

Progeny: Pp Pp pp pp

The birth of children is an independent event i.e. the first child will not impact the occurance of disease in the next child.

a) As shown in the cross above, the probability of next child being born with polydactyly is 50% or 1/2 because out of 4 children 2 have dominant allele and therefore will acquire the disease.

b) As already explained, birth of child is an independent event so the probability of their next 3 children being born as healthy will be 1/2 x 1/2 x 1/2 = 1/8 because probability of being born as normal is also 1/2.

c) In order to calculate probability in this situation binomial theoram can be used and the formula is as under:

Probability = n!/r! s! x (a)^r x (b)^s

where, n = Total no. of events

r = No. of times outcome 'a' occurred

s = No. of times outcome 'b' occurred

a = probability of 1st outcome

b = probability of 2nd outcome

Probability = 3!/2! 1! x (1/2)² x (1/2)¹

= 3 x 1/8

= 3/8