Two litters of a particular rodent species have been born, one 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.

(a) What is the probability that the animal chosen is brown-haired?
(b) Given that a brown-haired offspring was selected, what is the probability that the sampling was from litter 1?

Being that We are given,event of Brown hair with two disjoint events, one is { ( BrownHair ) ∩ ( Litter 1) } and the other is { ( BrownHair ) ∩ ( Litter 2) } .

a) To find the probability that the animal chosen is brown-haired,

Let A : we choose a brown-haired rodent and B : we choose litter1.

So using the axioms of probability, we can write

P(A) = P(A | B) * P(B) + P(A | Bc) * P(Bc)

Making use of the given information, we get;

number of brown haired rodents in litter 2 P(AB) Total number of rodents in litterl

and

P(A |B^{c}) =\frac{\text{number of brown haired rodents in litter2}}{\text{Total number of rodents in litter2}}= \frac{3}{5}

And also it is given that we choose litter at random ,so P(B) = P(Bc ) = 1/2

So now we plug these values in the equation of P(A) and get

P(A) = (\frac{2}{3}*\frac{1}{2}) + (\frac{3}{5}*\frac{1}{2}) = \frac{2}{6}+\frac{3}{10} = 0.633

So probability that the animal chosen is brown-haired = 0.633

b) Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A)

Lets make use of Bayes rule to find this conditional probability,

So using theorem we get,

P(B|A) = \frac{P(A|B)*P(B)}{P(A|B)*P(B)+P(A|B^{c})*P(B^{c})}

P(B|A) = \frac{(1/2)*(2/3)}{[(1/2)*(2/3)]+[(1/2)*(3/5)]} = \frac{10}{19} = 0.5263

Thus, Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A) = 0.5263.