Ok. So, to summarize the data:
Our prior (before we consider their offspring) is:
Mom is a healthy carrier with probability $\frac 23$, and a healthy non-carrier with probability $\frac 13$.
Dad is a healthy carrier with probability $\frac 2{21}$, and a healthy non-carrier with probability $\frac {19}{21}$.
Now they have two healthy kids. That will change our probability estimate (as it provides evidence that at least one parent is a healthy non-carrier).
Specifically: The prior probability that both parents are healthy carriers is $\frac 23\times \frac 2{21}=\frac 4{63}$. Conditioned on that, the probability that a given child has CF is $\frac 14$. Thus the probability, in this case, that both kids are healthy is $\left(\frac 34 \right)^2=\frac 9{16}$ Of course, if either parent is a non-carrier then the probability that a given child has CF is $0$. Thus our new estimate of the probability that both parents are carriers is $$\frac {\frac9{16}\times \frac 4{63}}{\frac 9{16}\times \frac 4{63}+1\times \frac {59}{63}}=.03673469$$
Note: that's considerably lower than our prior, which was $\frac 4{63}=.063492$.
Using this new estimate for the probability, we see that the probability that the next child has CF is $$\frac 14\times .03673469=.00918367$$ or about $.918\%$
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:
- 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!
Best Answer
If $\ \{F,M\}\ $ is the multiset of the genotypes of the brothers' parents, then $$ P(AH)=\sum_{g=G} P(AH|\{F,M\}=g)P(\{F,M\}=g)\ , $$ where $\ G\ $ is the set of all possible genotype pairs.
The possible genotype pairs of the parents are listed in the first column of the following table, the probabilities of those genotype pairs are listed in the second column, the conditional probabilities of the first son having genotype $\ Aa\ $ are listed in the third, and the conditional probabilities of both sons having genotype $\ Aa\ $ are listed in the fourth. Since the events that the sons have any particular genotype are conditionally independent given the parents' genotype pair, this latter conditional probability is just the square of the former. $$ \begin{array}{c|c|c|c|c|c|c|} g&P(\{F,M\}=g)&P(H|\{F,M\}=g)&P(AH|\{F,M\}=g)\\ \hline \{2AA\}&p^4&0&0\\ \{AA,Aa\}&4p^3q&\frac{1}{2}&\frac{1}{4}\\\ \{AA,aa\}&2p^2q^2&1&1\\ \{2Aa\}&4p^2q^2&\frac{1}{2}&\frac{1}{4}\\ \{Aa,aa\}&4pq^3&\frac{1}{2}&\frac{1}{4}\\ \{2aa\}&q^4&0&0\\ \hline \end{array} $$
Thus, \begin{align} P(AH)&=p^3q+3p^2q^2+pq^3\\ &=pq\big(p^2+3pq+q^2\big)\\ &=pq(1+pq)\ , \end{align} and therefore $$ P(AH|H)=\frac{1+pq}{2}\ . $$