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!

  • $\begingroup$ To be clear, there are $\theta$ men for every $1$ woman, correct? $\endgroup$ – Noble Mushtak Dec 28 '18 at 22:20
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    $\begingroup$ No, what I mean is P(A)/ P(A') = θ , where A is that the subject is male. (So I guess, men/women= θ) $\endgroup$ – butterflyflyaway Dec 28 '18 at 22:22
  • $\begingroup$ Condition the complement of your event on the number of females sampled: 0, 1, or 2. $\endgroup$ – Laars Helenius Dec 28 '18 at 22:26
  • $\begingroup$ Why only on females though? $\endgroup$ – butterflyflyaway Dec 28 '18 at 22:27

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 $P(A^C)=1-P(A)$:


Now, the probability that at least one person is colorblind has three cases in it:

  • Both people are colorblind -> $P(colorblind)^2$
  • First person is colorblind, second person is not -> $P(colorblind)\cdot P(colorblind^C)$
  • Second person is colorblind, first person is not -> $P(colorblind)\cdot P(colorblind^C)$

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!

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    $\begingroup$ To complete the problem, assume the sampling is independent. Then $P(\text{at least one colorblind})=1-P(\text{not colorblind})^2=1-[1-P(\text{colorblind})]^2$ $\endgroup$ – M. Nestor Dec 28 '18 at 22:31
  • $\begingroup$ One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people? $\endgroup$ – butterflyflyaway Dec 28 '18 at 22:33
  • $\begingroup$ @butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind. $\endgroup$ – Noble Mushtak Dec 28 '18 at 22:36
  • $\begingroup$ @M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did. $\endgroup$ – Noble Mushtak Dec 28 '18 at 22:38

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