Question

Two litters of a particular rodent species have been born; first with two brownhaired and one gray-haired (litter 1) and the other with three brown-haired and two gray-haired (litter 2). We select a litter at random and then select an offspring at random from the selected litter. (i) What is the probability that the animal chosen is brown-haired

Answer 1

Answer:

The probability that the animal chosen is brown-haired is 0.6333.

Step-by-step explanation:

Denote the events as follows:

A : a brown-haired rodent

B : Litter 1

The information provided is:

$P (A|B) =\frac{2}{3}\\\\P(A|B^{c})=\frac{3}{5}$

The probability of selecting any of the two litters is equal, i.e.

$P(B)=P(B^{c})=\frac{1}{2}$

According to the law of total probability:

$P(X)=P(X|Y_{1})P(Y_{1})+P(X|Y_{2})P(Y_{2})+...+P(X|Y_{n})P(Y_{n})$

Compute the total probability of event A as follows:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

         $=[\frac{2}{3}\times\frac{1}{2}]+[\frac{3}{5}\times\frac{1}{2}]\\\\=\frac{1}{3}+\frac{3}{10}\\\\=\frac{10+9}{30}\\\\=\frac{19}{30}\\\\=0.6333$

Thus, the probability that the animal chosen is brown-haired is 0.6333.