Answer: The probability that individual 1 is heterozygous is 1.
The probability that individual 3 is heterozygous is 1.
The probability that individual 4 is affected is 0.
Explanation: Since the father was said to be affected, the genotype of the father will be homozygous recessive. Using (A) to represent the dominant allele and (a) to represent the recessive allele, we can say that the father's genotype is aa.
Since the genotype of the mother was not given, we assume that she has wild-type genotype which is homozygous dominant allele because no contrary information was given. We can then say that her genotype is AA.
A cross between aa and AA will produce offspring with genotype Aa only. Therefore, we can conclude that 1 - 4 have heterozygous alleles.
Check the attached image for better illustration.