This is a question from Sheldon Ross's book 8th ed (Chapter 3, Q 61) –
Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from
both parents will be albino. Persons having the gene pair A, a are normal in appearance and,
because they can pass on the trait to their offspring, are called carriers. Suppose that a normal
couple has two children, exactly one of whom is an albino. Suppose that the nonalbino child mates
with a person who is known to be a carrier for albinism.
(a) What is the probability that their first offspring is an albino?
(b) What is the conditional probability that their second offspring is an albino given that their firstborn is not?
My solution process (part (a))
The normal couple can be any one of $(Aa, Aa), (Aa, aA), (Aa, aa)$ or $(aA, aa)$. The first and second combination will give gene pairs as $(AA, Aa, aA, aa)$ and third and fourth will give $(aa, aa, Aa, Aa)$. The non albino will not be $aa$, so when this nonalbino child mates with another carrier (without loss of generality, lets assume $Aa$), I will get $(6+4)$x$4 = 40$ possible offspring gene pairs ($AA-12, aa-8$ and remaining $Aa$ or $aA$). So I have 8 outcomes for getting an albino child out of 40, which gives the probability as $1/5$.
Another solution
I got a solution to first part here. The answer say that the nonalbino child (who mates with another carrier) is the offspring of two people having gene pairs $A,a$. I don't understand why they both have to be the carriers? If one of the parent is nonalbino (having $A,a$ or $A,A$ or $a,A$) and other is albino (having $a,a$), then also they can have one albino $(a,a)$ and one non albino $(A,A$ or $A,a)$ child. Is there something wrong with my reasoning?
Best Answer
We know both parents of the non-albino child must be carriers because they produced an $a,a$ offspring. The possibilities for the non-albino child are therefore, $$A,a$$ $$a,A$$ $$A,A$$
This means there is a $\frac{2}{3}$ probability that the non-albino child is a carrier. Each of the non-albino and his spouse donates an allele independently so we are looking for the product of the probabilities that each donates an $a$.
$P(\text{non-albino donates a})=P(\text{he is a carrier})P(\text{donates a}|\text{he is a carrier})=\frac{2}{3}\frac{1}{2}=\frac{1}{3}$
$P(\text{spouse donates a})=\frac{1}{2}$ because we know she is a carrier.
Therefore, the answer to the first question is $\frac{1}{3}\frac{1}{2}=\frac{1}{6}$
To get the second part you can use Bayes' theorem to compute $P(\text{second is}|\text{first is not})$ and by noticing that $P(\text{first is not})=\frac{5}{6}$