# problem colour of eyes

The color of a person’s eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent.

Suppose that Smith and both of his parents have brown eyes, but Smith’s sister has blue eyes. (a) What is the probability that Smith possesses a blue-eyed gene?

if Smith's sister has blue eyes , Smith's parents have both the genes Blue-Brown.

The possible cases for Smith are: Brown-Brown, Blue-Brown, Brown-Blue and the positive cases: Blue-Brown, Brown-Blue

$$p=2/3$$

(b) Suppose that Smith’s wife has blue eyes. What is the probability that their first child will have blue eyes?

$$p(\text{the first child has blue eyes})=p(\mathrm{B|Brown-Brown})*p(\mathrm{Brown-Brown})+p(\mathrm{B|Blue-Blue})*p(\mathrm{Blue-Blue})+p(\mathrm{B|Blue-Brown})*p(\mathrm{Blue-Brown})+p(\mathrm{B|Brown-Blue})*p(\mathrm{Brown-Blue})=0*\frac{1}3+1*0+\frac{1}2*\frac{1}3+\frac{1}2*\frac{1}3=\frac{1}3$$

(c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

C1=event "the first child has brown eyes", $p(C1)=2/3$

C2=event "the second child has brown eyes"

could someone give me a suggestion?

• Just to add some terminology. Given the statement of the problem, there would be a single gene for eye colour (which I think is not actually the case). However, the actual DNA sequence influencing if the eye colour is blue or brown is known as the allele: so in this problem there are blue and brown alleles. Think of the gene as the variable, and the allele as the value of the variable. Since you have two copies of the gene, the combination of the two alleles is known as the genotype, while the resulting eye colour is the phenotype. – Einar Rødland Sep 17 '17 at 10:50

Either Smith is BB (Brown-Brown) or Bb (Brown-blue); as you correctly calculated, $P(BB) = \frac13$.

The fact that the first child has brown eyes lessens the likelihood that Smith is Bb. Let $Brown$ represent the event that Smith's first child has brown eyes (given only the information knwon before the birth). Then
$$P(Brown\wedge BB) = 1\cdot \frac13 = \frac13 \\ P(Brown|\wedge Bb) = \frac12 \cdot \frac23 = \frac13\\ P(Brown) = \frac13 + \frac13 = \frac23\\ B(BB | Brown) = \frac13/\frac23=\frac12$$ So the probability of BB, now that we know that $Brown$ is true, is $\frac12$ and the probability of the next child having brown eyes is $$\frac12\cdot 1 + \frac12 \cdot \frac12 = \frac34$$

• excuse me what do you mean with BB? – Anne Apr 3 '17 at 18:02
• the final result is correct but I don't understand your solution. Can you be more clear? – Anne Apr 4 '17 at 7:24
• @Anne $BB$ means that Smith has two alleles for brown eyes, while $Bb$ means that Smith has one allele for brown eyes and one for blue eyes. – N. F. Taussig Sep 17 '17 at 10:39

Both answers, by Mark and foo, seem correct, but there's a slightly different way of explaining it (although not fundamentally different).

Like Mark, I let B and b denote the brown and blue alleles, respectively. I also let Brown and Blue denote the resulting eye colour: so bb results in Blue, while BB and Bb (same as bB if we don't care about the order) result in Brown.

As is already known, $\Pr[\text{Smith}=BB]=1/3$ and $\Pr[\text{Smith}=Bb]=2/3$. His wife is bb, so what determines the children's eye colour is which allele they inherit from Smith.

As has already been shown, the chance that the first child inherits the B allele from Smith is $1/3$: $$\Pr[C_1\sim\textit{Brown}] = \Pr[\text{Smith}=BB]\cdot\frac{2}{2} + \Pr[\text{Smith}=Bb]\cdot\frac{1}{2} = \frac{1}{3} + \frac{2}{3}\cdot\frac{1}{2} = \frac{2}{3}$$ where the first $1/2$ is the likelihood that the child inherits the B allele from Smith if he has Bb; and the $2/2$ just indicates that when Smith has both B alleles, the child will always inherit B.

We can make the same calculation asking for the likelihood that the first two children both have brown eyes: $$\Pr[C_1,C_2\sim\textit{Brown}] = \Pr[\text{Smith}=BB] + \Pr[\text{Smith}=Bb]\cdot\frac{1}{2^2} = \frac{1}{3} + \frac{2}{3}\cdot\frac{1}{4} = \frac{1}{2}.$$

This makes the conditional likelihood of the second child having brown eyes given that the first has brown eyes $$\Pr[C_2\sim \textit{Brown}\mid C_1\sim\textit{Brown}] = \frac{\Pr[C_1,C_2\sim\textit{Brown}]}{\Pr[C_1\sim\textit{Brown}]} = \frac{1/2}{2/3} = \frac{3}{4}.$$