Step 1: Understanding the Concept
Colour blindness is X-linked recessive.
Colourblind male is \(X^cY\) and a normal visioned female is \(XX\).
Step 2: Cross
Gametes of father: \(X^c\) and \(Y\).
Gametes of mother: \(X\) and \(X\).
Offspring: \(XX^c\) (carrier daughters) and \(XY\) (normal sons).
Step 3: Result
Every son gets normal X from the mother, so the probability of a colourblind son is zero.
Final Answer:
The probability of a colourblind son is zero.
\[ \boxed{\text{(A) }\text{zero}} \]