5% of men and 0.25% of women are colour blind, and men and women each make up half the population. A person is chosen at random. (1) What is the probability the person is colour blind? (2) Given that the person is colour blind, what is the probability the person is male?

Step-by-step solution

1. 1

   Set up the notation and the given probabilities. Let $M$ = male, $W$ = female, $B$ = colour blind. The data are

   $P(M)=P(W)=0.5,\qquad P(B\mid M)=0.05,\qquad P(B\mid W)=0.0025$

   Note $0.25\%=0.0025$, not $0.025$ — a factor-of-ten slip here changes both answers.

2. 2

   (1) Apply the law of total probability. $M$ and $W$ partition the population, so

   $P(B)=P(M)P(B\mid M)+P(W)P(B\mid W)$

3. 3

   Evaluate part (1).

   $P(B)=0.5(0.05)+0.5(0.0025)=0.025+0.00125=0.02625$

   So about $2.6\%$ of the population is colour blind — sensibly between the male rate $5\%$ and the female rate $0.25\%$.

4. 4

   (2) Apply Bayes' theorem. The question reverses the conditioning:

   $P(M\mid B)=\frac{P(M)P(B\mid M)}{P(B)}=\frac{0.025}{0.02625}$

5. 5

   Evaluate part (2). Multiplying numerator and denominator by $10000$ and cancelling,

   $\frac{250}{262.5}=\frac{2000}{2100}=\frac{20}{21}\approx 0.9524$

   $\boxed{P(B)=0.02625,\qquad P(M\mid B)=\tfrac{20}{21}\approx 0.952}$

6. 6

   Sanity-check the reversal. Even though only half the population is male, $95\%$ of colour-blind people are male — because the male rate is $20$ times the female rate. Confirming the complement: $P(W\mid B)=\tfrac{0.00125}{0.02625}=\tfrac{1}{21}$, and $\tfrac{20}{21}+\tfrac{1}{21}=1$ ✓.