Q.3.15 - A First Course in Probability

Question

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? 

Step-by-Step Solution

Answer

a). $\frac{1}{3}\Rightarrow$ Figure out the parent's genes, each of the possible gene pairs (mothers gene, fathers gene) has the same probability of occurring.

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

Step 1: Given Information (Part a)

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

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

Step 2: Explanation (Part a)

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

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

$P\left[\right(B,B\left)\right]=\frac{1}{3}$.

Step 3: Final Answer

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

Step 4: Given Information (Part b)

b - gene for brown color (not dominant).

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

Step 5: Explanation (Part b)

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 $A^c=\left\{\right(B,b),(b,B\left)\right\}$

$P[A\mid 5\text{ children black }]=\frac{P[5\text{ children black }\mid A]P\left(A\right)}{P[\text{ 5 children black }\mid A]P\left(A\right)+P\left[5\text{ children black }\mid A^c\right]P\left(A^c\right)}$

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

$P[\text{ all }5\text{ children are black }\mid A]=P[1\text{ child is black }\mid A{]}^5$

Step 6: Explanation (Part b)

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 $(b, b)$

$P[1 child is black | A]=1$

$P[1 child is black | A^c]=\frac{1}{2}$

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

$P[A\mid \text{ 5 children black }]=\frac{P[1\text{ child is black }\mid A{]}^{5}P(A)}{P[1\text{ child is black }\mid A{]}^{5}P(A)+P{\left[1\text{ child is black }\mid A^c\right]}^{5}P\left(A^c\right)}$

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

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

Step 7: Final Answer (Part b)

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