Question

Suppose 5 men out of 100 and 25 women out of 1000 are colour blind. a colour blind person is chosen at random. what is the probability of his being male (assuming that male and females are equal in proportion).

Answer 1

total persons = 100 +1000
                     = 1100

total outcome = 1100

colour blind man = 5

probability of getting a blind man  = 5/1100
=0.0045

Answer 2

Answer:

The probability of his being male is $\frac{20}{21}$

Step-by-step explanation:

Given : Suppose 5 men out of 100 and 25 women out of 10000 are colour blind. A colour blind person is chosen at random.

To find : The probability of this being a male is equal to (Assume males and females to be equal in number) ?

Solution :

Let $A_1$ be the event of selection a male.

Let $A_2$ be the event of selection a female.

$P(A_1)=P(A_2)=\frac{1}{2}$

Let A be the event of selection colour blind person

$P(A/A_1)=\frac{5}{100}=\frac{1}{20}$

$P(A/A_2)=\frac{25}{10000}=\frac{1}{400}$

Applying Bayes theorem,

$P(A_1/A)=\frac{P(A_1)P(A/A_1)}{P(A_1)P(A/A_1)+P(A_2)P(A/A_2)}$

$P(A_1/A)=\frac{\frac{1}{2}\times \frac{1}{20}}{\frac{1}{2}\times \frac{1}{20}+\frac{1}{2}\times \frac{1}{400}}$

$P(A_1/A)=\frac{\frac{1}{40}}{\frac{1}{40}+\frac{1}{800}}$

$P(A_1/A)=\frac{\frac{1}{40}}{\frac{20+1}{800}}$

$P(A_1/A)=\frac{\frac{1}{40}}{\frac{21}{800}}$

$P(A_1/A)=\frac{1}{40}\times \frac{800}{21}$

$P(A_1/A)=\frac{20}{21}$

Therefore, The probability of his being male is $\frac{20}{21}$