Question

Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both parents will be albino. Persons having the gene pair A, a are normal in appearance and, because they can pass on the trait to their offspring, are called carriers. Suppose that a normal couple has two children, exactly one of whom is an albino. Suppose that the non albino child mates with a person who is known to be a carrier for albinism.

(a) What is the probability that their first offspring is an albino?

(b) What is the conditional probability that their second offspring is an albino given that their firstborn is not?

Short answer

1. The probability that their first offspring is an albino is $\frac{1}{6}$.
2. The conditional probability that their second off spring is an albino given that their first is$\frac{3}{20}$
   

Step-by-step solution

Given information (Part a)

Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both parents will be albino.

Solution (Part a)

The following information is provided.

The individuals who acquire the 'a' gene from both parents will be albino. The individual with gene pair A, a is in ordinary formation and called a carrier. Genes correlating albinism are represented by ' $A^{\prime}$ and ' a '.

A Regular couple has two kids one is with albino and the other doesn't have albino. The nonalbino child mates with another person who is known to be normal.

An albino person with gene pair (a, a) and nonalbino individual with gene pairs are (A, A),(a, A)

Here the non-albino kid mates with an individual known to be an albino carrier.

A non-albino kid is an individual understood to be a carrier or not a carrier. If an individual with the gene pairs (a, A) or (A, a) is called a carrier and with the gene pair (A, A) is a non-carrier.

The probability of albino child is $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

The probability that their first offspring is an albino can be defined as follows.

The probability that the first offspring is an albino and non-albino child is a carrier + Probability that the first offspring is an albino and non-albino child is not a carrier.

Therefore the probability that their first off-spring is an albino

=P( NA child is carrier) P( off spring is albino | non albino child is carrier)+

P (NA child is not carrier) P (off spring is albino| non albino child is a not a carrier)

(NA means Non albino)

$=\left(\frac{2}{3}\right)\left(\frac{1}{4}\right)+\left(\frac{1}{3}\right)\left(0\right)$

$=\frac{1}{6}$

Final answer (Part a)

The probability that their first off-spring is an albino$=\frac{1}{6}$.

Given information (Part b)

Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both parents will be albino.

Solution (Part b)

Probability that the first off-spring is not an albino is

$=1-P\left(\text{First off spring is an albino}\right)$

$=1-\frac{1}{6}$

$=\frac{5}{6}$

Suppose O₁ be the occasion that the first offspring is an albino and O₂ be the second offspring is an albino and C is non-albino child is a carrier.

The conditional probability that their second offspring is an albino given that their first is not is as follows.

$=P\left(O_2\mid O_1^c\right)$

$=\frac{P\left(O_2{O}_1^c\right)}{P\left(O_1^c\right)}$

$=\frac{P\left(C\right)P\left(O_2{O}_1^c\mid C\right)+P\left(C^c\right)P\left(O_2{O}_1^c\mid C^c\right)}{P\left(\text{lst off spring is not}\right)}$

$=\frac{(2/3)(3/4)(1/4)+(1/3)\left(0\right)}{5/6}$

$=\frac{3}{20}$

Final answer (Part b)

The conditional probability that their second off spring is an albino given that their first is$\frac{3}{20}$.