According to statistics, 5% of men and 0.25% of women are color blind. Before obtaining a driver’s license

According to statistics, 5% of men and 0.25% of women are color blind. Before obtaining a driver’s license, a group of 4 women and 11 men is examined. A person with color blindness is chosen at random. What is the likelihood that this is a woman?

We denote by Hm; Hj – hypothesis that the chosen one is a man or a woman.
1. Let’s find the probability of hypotheses Hm; Hzh;
The probabilities of these hypotheses before testing are:
P (Hm) = 1/11 = 0.091;
P (Hw) = 1/4 = 0.25;
2. Find conditional probabilities;
Event A – Colorblind Chosen. Let us find the conditional probabilities of this event under hypotheses Hm; Hzh;
Color blind man:
P (A | Hm) = m / n = 5/100 = 0.05;
Color blind woman:
P (A | Hl) = m / n = 0.25 / 100 = 0.0025;
Let’s apply Bayes’ formula:
The likelihood that a color blind man is:
P (Hm | A) = (P (A | Hm) P (Hm)) / (P (A | Hm) P (Hm) + P (A | Hm) P (Hm)) =
= (0.05 0.091) / (0.05 0.091 + 0.0025 0.25) = 0.00455 / 0.005175 = 0.879;
The likelihood that a woman is color blind:
P (Hl | A) = (P (A | Hl) P (Hl)) / (P (A | Hl) P (Hl) + P (A | Hm) P (Hm)) =
= (0.0025 0.25) / (0.0025 0.25 + 0.05 0.091) = 0.000625 / 0.005175 = 0.121;
Answer: The probability that a color-blind person is a man P (Hm | A) = 0.879, a woman P (Hw | A) = 0.121;