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?

a. 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) = 0.5263.

Step-by-step explanation:

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) = / fracP (A{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.