I have a statistics problem and, although I have solved most of it, I am stuck for quite some time in the last question.
Say that there we have a population where the analogy of men to women is θ. Suppose that the probability of a woman being colorblind is $q^2$ and the probability of a man being colorblinf $q$.
Up to now I have found, if choosing one random person, the probability of him/her being colorblind and also that probability of a person being a man, given that he is colorblind. However, now I can't find a way to approach this question:
Suppose we choose two people out of the population radomly. What is the probability that at least one person is colorblind?
I was thinking of approaching it by finding the probability of none being colorblind and then finding its compliment, but I can't figure out how to do this, given that we have a different percentage for men and women. Any help on this would be appreciated.
Thank you!
Best Answer
Let there be $m$ men and $w$ women (unfortunately, I also have to assume there are also $0$ non-binary people in this population). We have the following:
$$\frac m w=\theta\rightarrow m=\theta w$$
Thus, the probabilities of being a man or a woman are:
$$P(man)=\frac{m}{m+w}=\frac{\theta w}{\theta w+w}=\frac{\theta}{\theta+1}$$ $$P(woman)=\frac{w}{m+w}=\frac{w}{\theta w+w}=\frac{1}{\theta+1}$$
Now, we are also given the following:
$$P(colorblind | man)=q^2$$ $$P(colorblind | woman)=q$$
Then, we can use the formula $P(A \cap B)=P(B)P(A|B)$ to find the following:
$$P(colorblind \cap man)=P(man)P(colorblind | man)=\frac{q^2\theta}{\theta+1}$$ $$P(colorblind \cap woman)=P(woman)P(colorblind | woman)=\frac{q}{\theta+1}$$
Finally, since man and woman are the only two cases in this population, $woman=man^C$—that is, woman is the complement case of man. Thus, we can use the formula $P(A)=P(A \cap B)+P(A \cap B^C)$ to find:
$$P(colorblind)=P(colorblind \cap man)+P(colorblind \cap woman)=\frac{q^2\theta+q}{\theta+1}$$
And of course, to find the probability is not colorblind, we use $P(A^C)=1-P(A)$:
$$P(colorblind^C)=1-P(colorblind)=\frac{(1-q^2)\theta+1-q}{\theta+1}$$
Now, the probability that at least one person is colorblind has three cases in it:
Thus, to find the total probability of there being one color blind person, we just add up all of these cases:
$$P(at\ least\ one\ colorblind)=P(colorblind)^2+2P(colorblind)\cdot P(colorblind^C) \\=\frac{(q^2\theta+q)^2+2(q^2\theta+q)(\theta(1-q^2)+1-q)}{(\theta+1)^2}$$
I'll leave doing the algebra to you. Good luck!