91Ó°ÊÓ

Given that 5 percent of men and 0.25 percent of women are color blind. A color-blind person is chosen at random

Also given that there are an equal number of males and females.

So, the probability for males is $0.5$and

probability for a female is$0.5$

We have to find he conditional probability that a person is male, given that he has colorblind

Diagram Tree

Let $E$denote the event that the person has colorblind$,F$denote the event that the selected person is female, and$M$denote the event that the selected person is male.

Thus,

$P(Mâˆ\pounds E)=0.05$, $P(Fâˆ\pounds E)=0.0025,\text{and}P\left(M\right)=P\left(F\right)=0.5$

The conditional probability that a person is male, given that he has colorblind is,

$P(Mâˆ\pounds C)=\frac{P(Câˆ\pounds M)P\left(M\right)}{P(Câˆ\pounds M)P\left(M\right)+P(Câˆ\pounds F)P\left(F\right)}$

$=\frac{0.05×0.5}{(0.05×0.5)+(0.0025×0.5)}$

$=\frac{0.025}{0.02625}$

$=0.9524$

The conditional probability that a person is male, given that he has colorblind is 0.9524

Given that 5 percent of men and 0.25 percent of women are color blind. A color-blind person is chosen at random

Also given that there are an equal number of males and females.

So, the probability for males is 0.5 and probability for a female is0.5

We have to find the conditional probability that a person is male, given that he has colorblind

Suppose the population consisted of twice as many males as females.

Then, the tree diagram is shown in below:

So,

$P(Mâˆ\pounds E)=0.05,P(Fâˆ\pounds E)=0.0025,P\left(M\right)=\frac{2}{3},\text{and}P\left(F\right)=\frac{1}{3}$

The conditional probability that person is male, given that he has colorblind is,

$P(Mâˆ\pounds C)=\frac{P(Câˆ\pounds M)P\left(M\right)}{P(Câˆ\pounds M)P\left(M\right)+P(Câˆ\pounds F)P\left(F\right)}$

$=\frac{0.05×\frac{2}{3}}{\left(0.05×\frac{2}{3}\right)+\left(0.0025×\frac{1}{3}\right)}$

$=\frac{0.033333}{0.034167}$

=0.9756

The conditional probability that a person is male, given that he has colorblind is0.9756