Since each child of two heterozygous parents has a 50% chance of receiving a recessive trait from each parent,
a. if the first child is phenotypically recessive, then the next child must be phenotypically dominant.
b. if the first child is phenotypically recessive, then the next child has a 3/4 chance of being phenotypically recessive.
c. if the first child is phenotypically recessive, then the next child has a 1/2 chance of being phenotypically recessive.
d. no matter what the first child's phenotype, the next child will have a 1/4 chance of being phenotypically recessive.

d. no matter what the first child's phenotype, the next child will have a 1/4 chance of being phenotypically recessive.

Explanation:

The birth of any child with a particular trait is an independent event which means that it does not matter what is the phenotype/genotype of first child, it will not impact the phenotype/genotype of the next child.

In general, each child of two heterozygous parents has a 50% chance of receiving a recessive trait from each parent. Then in such scenario, the probability of a child to be born with a recessive phenotype is 1/4.

Let us consider a trait with alleles R and r. The cross is depicted in the picture attached. It is clear from the picture that out of the 4 probable progeny, 3 with 'genotype RR, Rr and Rr' will show dominant trait phenotypically while only 1 with 'genotype rr' will show recessive trait. It means that chance of dominant progeny is 3/4 while that of recessive progeny is 1/4.

But, birth of the child is independent event so if first child is born with recessive trait, the next child will still have 1/4 chance of having recessive trait phenotypically .

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