In a certain species of rats, black dominates over brown. Suppose that a black rat with two black parents has a brown sibling.

(a) What is the probability that this rat is a pure black rat (as opposed to being a hybrid with one black and one brown gene)?

(b) Suppose that when the black rat is mated with a brown rat, all$5$ of their offspring are black. Now what is the probability that the rat is a pure black rat?

$B$- gene for black color (dominant).

Black rat in question, black $\to (B,B),(b,B),(B,b)$.

$P\left[\right(B,B\left)\right]=?$

The requested probability is that the rat in question has $(B,B)$ genes.

From the note that all gene combinations are equally possible, and there are three of them

$\frac{1}{3}\Rightarrow$ Figure out the parents genes, each of the possible gene pairs (mothers gene, fathers gene) has the same probability of occuring.

b - gene for brown color (not dominant).

A person has two genes for eye-color - $(m,f)$.

Rat's mate is brown, $P\left[\right(BB)\mid$ all $5$ children black $]=$ ?

Bayes formula with system of events being $A=(B,B)$ and $A^c=\left\{\right(B,b),(b,B\left)\right\}$

Given the genes of the rat in question, the color of the children are independent, in this notation:

Same for $A^c$.

Taking into account genes $A=(B,B)$and $A^c=\left\{\right(B,b),(b,B\left)\right\}$, and the mates genes are $(b,b)$

$P[1childisblack|A]=1$

$P[1childisblack|A^c]=\frac{1}{2}$

And from a) $P\left(A\right)=\frac{1}{3},P\left(A^c\right)=\frac{2}{3}$thus the Bayes formula is:

$=\frac{1^5·\frac{1}{3}}{1^5·\frac{1}{3}+{\left(\frac{1}{2}\right)}^5·\frac{2}{3}}$

$=\frac{2^4}{2^4+1}$

$\frac{2^4}{2^4+1}\Rightarrow$ Bayes formula with conditioning on genes of that rat.