Help a misunderstood Bayesian mathematician

Question

I'm a mathematician-turned-physician and I'm currently taking a Genetics course. Some high-school level probability is usually covered in these courses and it would be expected that lecturers that are not mathematicians/statisticians may not fully comprehend what they are talking about.

The lecturer responsible for writing and grading exams made an inaccurate assertion and when I tried to make my case, she doubled down, claiming the answer depends on how ones sees it. Not the reply that a mathematician would like to hear but here we are. The problem is (sorry if I'm defining very simple terms/concepts):

It is given that an individual can have 4 equiprobable genotypes: $AA$, $Aa$, $aA$ and $aa$, where $A$ is the good/normal allele and $a$ is the defective allele. For completeness, we all carry 2 copies of the same gene/allele for all autosomal chromosomes (one we inherited from mom and the other from dad). Considering an autosomal recessive condition, i.e., the individual has the condition when both alleles are defective -- the $aa$ genotype -- what is the probability that the individual is a carrier, i.e., it would be possible for the individual to pass the defective allele forward to her progeny -- given that she is not affected, i.e., she does not have the $aa$ genotype. Moreover, do we need to apply Bayes' Theorem somehow to arrive at this conclusion (even if it is a very straightforward application)?

Edit: It is clear that the answer is $\frac{2}{3}$. This is not in contention. She claimed that "we did not need Bayes' Theorem" or any knowledge of conditional probability whatsoever to arrive at that conclusion. This is false as the moment we are restricting the sample space we are implicitly calculating a conditional probability, which would imply Bayesian reasoning, even if we do it without noticing it. It is a subtle point but it is an important one. I wondered whether other mathematicians/statisticians would agree with me... Formal solution below.

Answer 1

According to your problem statement, AA, Aa, aA and aa are equiprobable a priori, -- i.e. relative likelihood 1 : 1 : 1 : 1. Given the new information that "the person is not affected -- i.e. homozygous recessive aa", we now have relative likelihoods 1 : 1 : 1 : 0.

So AA, Aa, aA are equally likely and there is a $2/3$ chance that the person is a carrier.

Answer 2

Here's the formal Bayesian solution (unnecessary, really, because the problem is very simple.)

Let $C$ correspond to the event that the individual is a carrier, i.e., she carries at least one deleterious allele and let $H$ correspond to the event that the individual is healthy, i.e., she does not have the disease. We are interested in calculating the probability $P(C|H)$. Applying Bayes' Theorem: \begin{eqnarray*} P(C|H) & = & \frac{P(C \cap H)}{P(H)} \\ & = & \frac{P(H|C) P(C)}{P(H)} \\ & = & \frac{P(H|C) P(C)}{P(H|C) P(C) + P(H|\bar{C}) P(\bar{C})} \end{eqnarray*} where $\bar{C}$ corresponds to the event that the individual does not harbor any deleterious allele.

The unconditional probabilities $P(C)$ and $P(\bar{C})$ are trivial: $P(C) = \frac{3}{4}$ and $P(\bar{C}) = \frac{1}{4}$. The conditional probabilities $P(H|C)$ and $P(H|\bar{C})$ are also trivial: $P(H|C) = \frac{2}{3}$ and $P(H|\bar{C}) = 1$. Plugging all the numbers:

\begin{eqnarray*} P(C|H) & = & \frac{P(H|C) P(C)}{P(H|C) P(C) + P(H|\bar{C}) P(\bar{C})} \\ & = & \frac{\frac{2}{3} \times \frac{3}{4}}{\frac{2}{3} \times \frac{3}{4} + 1 \times \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3} \end{eqnarray*}