Probability

Question

Suppose that a particular trait (such as eye color or left-handedness) of a person is classified on the basis of one pair of genes, and suppose that d represents a dominant gene and r a recessive gene. Thus, a person with dd genes is pure dominance, one with rr is pure recessive, and one with rd is hybrid. The pure dominance and the hybrid are alike in appearance. Children receive 1 gene from each parent. If, with respect to a particular trait, 2 hybrid parents have a total of 4 children, what is the probability that 3 of the 4 children have the outward appearance of the dominant gene?

Answer 1

A hybrid parent has rd gene pair and it can contribute either r or d to a child. Considering the scenario where both parents are hybrid i.e rd, both can contribute r or d. The resultant child can have gene pair among these (rr, dd, rd) with rr having probability 1/4, same goes for dd and rd having probability 1/2 as rd and dr would mean the same type of person.

Now to have an appearance similar to dominance type , child should either be dd or rd whose probability is $\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$  whereas probability of not looking like dominance type is 1/4 (pure recessive rr).

Probability that 3 of the children have the outward appearance of the dominant gene : $\frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{1}{4}$ * 4 (as any of the 4 children can be of rr type so permutation does not matter)

The value equals to : $\frac{27}{64}$   This is the required probability