The following is a rephrase attempt of a previously unclear post I've made. The following question is related entirely to calculations and probability, however, it does slightly include genetics. I've solved the first 2 parts, but I can't solve the 3rd. I'm including all my calculations of the first 2 parts as well.
1) 0.25% of the entire population suffers from cystic fibrosis (CF), an autosomally recessive hereditary disease. What's the rate of heterozygotes in the population?
According to Hardy-Weinberg equations $$p^2 + 2pq +q^2 = 1, p + q = 1$$
Where $q^2$ is the rate of the genotype $aa$, and therefore sick people, $2pq$ is the rate of the genotype $Aa$, and therefore heterozygote people, a.k.a carriers but healthy, and $p^2$ is the rate of the genotype $AA$ and therefore completely healthy and not carriers either.
I take $q^2=0.25\% = 0.0025$, which gives $q=\frac1{20}$ and then $p=\frac{19}{20}$. $2pq$, a.k.a the rate of heterozygotesis then $\frac{38}{400} = 0.095 = 9.5\%$.
2) A couple goes to a consulation, to learn about their chances of having healthy/sick children, because the mother has a brother who suffers from this disease. Her 2 parents are healthy, and she is healthy as well. The husband is healthy, and has no history of CF in the family. What is the chance of them having a sick child?
First, let's establish, that if both the husband and wife are healthy, then for their child to be sick it requires them both to be with a genotype of $Aa$
Starting with the mother. If the mother is healthy, and her brother is sick, and both her parents are healthy, then it requires both parents' genotype to be $Aa$. Given that this is their genotype, there is a probability of $\frac14$ for the mother to be $AA$, $\frac12$ to be $Aa$ and $\frac14$ to be $aa$, but because we know she's healthy, she isn't $aa$, and so the probabilities change to $\frac23$ healthy carrier with genotype $Aa$, and $\frac13$ to be a healthy-noncarrier with genotype $AA$.
Now let's move to the father. He is healthy. So we are asking what are the chances of him being a carrier (=heterozygote), out of the "carriers + noncarriers" population, a.k.a $\frac{2pq}{2pq+p^2}=\frac2{21}$.
In the final calculation, we require both parents to be carriers, and the child to randomally have the genotype of $aa$, which has the probability of $\frac14$: $P$(mother is carrier)$\cdot$ $P$(father is carrier)$\cdot$ $P$(child's genotype is aa)$= P$(child is sick) $=\frac23 \cdot \frac2{21} \cdot \frac14 = \frac1{63}=1.58\%$
3) Given the couple has 2 healthy children, what is the chance of them having their 3rd child be born sick?
This is what I am not able to solve. The answer is approximately $0.9\%$. The calculation involves Bayes's formula and conditional probability.
Many thanks in advance!
Best Answer
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 $.918\%$