A husband and wife have normal vision. Both his father and her father are red-green colour-blind.. What is the probability that their first child to be a:

(a) son with normal vision

(b) son with colour blindness

(c) daughter with normal vision

(d) daughter with colour blindness

	(a)	(b)	(c)	(d)
(1)	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{2}$	$0$
(2)	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	$0$
(3)	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$
(4)	$\frac{1}{2}$	$0$	$\frac{1}{4}$	$\frac{1}{4}$

The woman is a carrier since her father is colour-blind.

Half of the sons are colour-blind. Both normal daughters (X^CX^C) and carrier daughters (X^CX^c) have normal vision. None of the daughters will be colour-blind.

Probability of their first child to be a son is $\frac{1}{2}$.

Probability of their first child to be a daughter is $\frac{1}{2}$.

Probability of their first child to be a normal son is $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$.

Probability of their first child to be a colour-blind son is $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$.

Probability of their first child to be a daughter with normal vision is $\frac{1}{2}\times 1=\frac{1}{2}$.

Probability of their first child to be a colour-blind daughter  is $\frac{1}{2}\times 0=0$.