Let:

• \(M\) = event that a person is male, \(P(M) = 0.54\)

• \(F\) = event that a person is female, \(P(F) = 0.46\)

• \(C\) = event that a person is colour-blind

We know:

• \(P(C|M) = 0.05\)

• \(P(C|F) = 0.02\)

We need to find \(P(M|C)\), the probability that a person is male given that they are colour-blind.

Using Bayes' theorem:

\(P(M|C) = \frac{P(C|M) \cdot P(M)}{P(C)}\)

First, calculate \(P(C)\):

\(P(C) = P(C \cap M) + P(C \cap F)\)

\(P(C \cap M) = P(C|M) \cdot P(M) = 0.05 \times 0.54 = 0.027\)

\(P(C \cap F) = P(C|F) \cdot P(F) = 0.02 \times 0.46 = 0.0092\)

\(P(C) = 0.027 + 0.0092 = 0.0362\)

Now, substitute back into Bayes' theorem:

\(P(M|C) = \frac{0.027}{0.0362} \approx 0.746\)

Thus, the probability that the person is male given that they are colour-blind is approximately 0.746, or \(\frac{135}{181}\).