Question 73, ch 2, Blitzstein
Blitzstein's Intro to Probability provides the following exercise:
In humans (and many other organisms), genes come in pairs. A certain gene comes in two types (alleles): type a and type A. The genotype of a person for that gene is the types of the two genes in the pair: AA; Aa; or aa (aA is equivalent to Aa). Assume that the Hardy-Weinberg law applies here, which means that the frequencies of AA; Aa; aa in the population are $p^2$; $2p(1 - p)$; $(1 - p)^2 respectively, for some p with 0 < p < 1. When a woman and a man have a child, the child's gene pair has one gene contributed by each parent. Suppose that the mother is equally likely to contribute either of the two genes in her gene pair, and likewise for the father, independently. Also suppose that the genotypes of the parents are independent of each other (with probabilities given by the Hardy-Weinberg law).
(3) Suppose that having genotype aa results in a distinctive physical characteristic, so it is easy to tell by looking at someone whether or not they have that genotype. A mother and father, neither of whom are of type aa, have a child. The child is also not of type aa. Given this information, find the probability that the child is heterozygous.
Hint: Use the definition of conditional probability. Then expand both the numerator and the denominator using LOTP, conditioning on the genotypes of the parents.
My issue: I know how to solve this question following the hint provided (see below) and with an alternative solution that is based on a substantive interpretation of the problem. However, my issue is that these two ways of solving the problem are not matching yet both ways of solving them seems logical. I am therefore hoping that you can discover what I am missing.
Definitions: $C_{a a}$ is the event that a child has gene combinations aa. $F_{a a}$ is the event that a father has gene combinations aa. $M_{a a}$ is the event that a mother has gene combinations aa. Let $N=C_{A a} \cap M_{a a}^{c} \cap F_{a a}^{c} \text { and } D=C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c}$
The aim is to find $P(C_{A a}| C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c} )$.
The first way of solving the problem (use the definition of conditional probability):
$P\left(C_{A a} | C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c}\right)=\frac{P(N)}{P(D)}$
We start with calculating the numerator: $$ \begin{aligned} P(G)=& P\left(G | M_{A A}, F_{A A}\right) P\left(M_{A A}, F_{A A}\right)+P\left(G | M_{A A}, F_{A a}\right) P\left(M_{A A}, F_{A a}\right) \\&+ P\left(G | M_{A a}, F_{A A}\right) P\left(M_{A a}, F_{A A}\right)+P\left(G | M_{A a}, F_{A a}\right) P\left(M_{A a}, F_{A a}\right) \\=&0 +\frac{1}{2} p^{2}(2 p(1-p))+\frac{1}{2} p^{2}(2 p(1-p))+\frac{1}{2}(2 p(1-p))^{2} \\=& 2 p^{2}(1-p). \end{aligned} $$
Then calculate the denominator: $$ \begin{aligned} P(H)=& P\left(H | M_{A A}, F_{A A}\right) P\left(M_{A A}, F_{A A}\right)+P\left(H | M_{A A}, F_{A a}\right) P\left(M_{A A}, F_{A a}\right) \\ &+P\left(H | M_{A a}, F_{A A}\right) P\left(M_{A a}, F_{A A}\right)+P\left(H | M_{A a}, F_{A a}\right) P\left(M_{A a}, F_{A a}\right) \\ =& p^{4}+p^{2}(2 p(1-p))+p^{2}(2 p(1-p))+\frac{3}{4}(2 p(1-p))^{2} \\ =& p^{2}(3-2 p) \\ \end{aligned} $$
Add these two facts together and we get: $P\left(C_{A a} | C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c}\right) =\frac{2 p^{2}(1-p)}{p^{2}(3-2 p)}$
The second way of solving the problem (expand the "conditioning" and intepret it substantively):
Begin with expanding the conditioning statments and use the fact that certain events are mutually exclusive (e.g. $C_{aa}^c=C_{A a} \cup C_{AA}=C_{A a} + C_{AA}$)
$$ \begin{aligned} D=C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c}=& (C_{A a} \cup C_{AA}) \cap (M_{A a} \cup M_{AA}) \cap (F_{A a} \cup F_{AA}) \\ =& (C_{A a} \cap M_{A a} \cap F_{AA}) + (C_{A a} \cap M_{A A} \cap F_{AA}) + (C_{A a} \cap M_{A a} \cap F_{Aa}) \\ &+ (C_{A a} \cap M_{A A} \cap F_{Aa}) + \\ & (C_{A A} \cap M_{A a} \cap F_{AA}) + (C_{A A} \cap M_{A A} \cap F_{AA}) + (C_{A A} \cap M_{A a} \cap F_{Aa}) \\ &+ (C_{A A} \cap M_{A A} \cap F_{Aa}) . \\ \end{aligned} $$
This decomposition of the conditioning term implies that we can decompose our quantity of interest into the following statment weighted by their probability of occuring,
$$ \begin{aligned} P(C_{A a}| D) =& P(C_{A a}| C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c} )= \\ =&P(C_{A a}|(C_{A a} \cup C_{AA}) \cap (M_{A a} \cup M_{AA}) \cap (F_{A a} \cup F_{AA}) ) \\ =& P(C_{A a}|C_{A a} \cap M_{A a} \cap F_{AA})P(C_{A a} \cap M_{A a} \cap F_{AA}) \\ &+ P(C_{A a}|C_{A a} \cap M_{A A} \cap F_{AA}) P( C_{A a} \cap M_{A A} \cap F_{AA})\\ &+ P(C_{A a}|C_{A a} \cap M_{A a} \cap F_{Aa}) P( C_{A a} \cap M_{A a} \cap F_{Aa})\\ &+ P(C_{A a}|C_{A a} \cap M_{A A} \cap F_{Aa}) P( C_{A a} \cap M_{A A} \cap F_{Aa}) \\ & + P(C_{A a}|C_{A A} \cap M_{A a} \cap F_{AA}) P(C_{A A} \cap M_{A a} \cap F_{AA})\\ &+ P(C_{A a}|C_{A A} \cap M_{A A} \cap F_{AA}) P(C_{A A} \cap M_{A A} \cap F_{AA} )\\ &+ P(C_{A a}|C_{A A} \cap M_{A a} \cap F_{Aa}) P( C_{A A} \cap M_{A a} \cap F_{Aa})\\ &+ P(C_{A a}|C_{A A} \cap M_{A A} \cap F_{Aa}) P( C_{A A} \cap M_{A A} \cap F_{Aa}) \\ \end{aligned} $$
From here we conclude that all statments that contain events with the child being of different genes on both sides of the conditioning is zero since they are mutually exclusive i.e. $P(C_{A a}|C_{A A} ,...)=0$. The opposite is true for all $P(C_{A a}|C_{A a} ,...)=1$ except for $P(C_{A a}|C_{A a} \cap M_{A A} \cap F_{AA})=0$ becuse the parents are all of AA genes and thus canno produce Aa children.
This fact simplifies the previous statment to the following,
$$ \begin{aligned} P(C_{A a}| D) =& P(C_{A a}| C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c} )= \\ =&P(C_{A a}|(C_{A a} \cup C_{AA}) \cap (M_{A a} \cup M_{AA}) \cap (F_{A a} \cup F_{AA}) ) \\ =& 1 P(C_{A a} \cap M_{A a} \cap F_{AA}) \\ &+ 0 P( C_{A a} \cap M_{A A} \cap F_{AA})\\ &+ 1 P( C_{A a} \cap M_{A a} \cap F_{Aa})\\ &+ 1 P( C_{A a} \cap M_{A A} \cap F_{Aa}) \\ & + 0P(C_{A A} \cap M_{A a} \cap F_{AA})\\ &+ 0P(C_{A A} \cap M_{A A} \cap F_{AA} )\\ &+ 0P( C_{A A} \cap M_{A a} \cap F_{Aa})\\ &+ 0 P( C_{A A} \cap M_{A A} \cap F_{Aa}) \\ =& P(C_{A a} \cap M_{A a} \cap F_{AA}) \\ &+ P( C_{A a} \cap M_{A a} \cap F_{Aa})\\ &+ P( C_{A A} \cap M_{A A} \cap F_{Aa}) \\ =& P(C_{A a} | M_{A a} \cap F_{AA}) P(M_{A a} \cap F_{AA}) \\ &+ P( C_{A a}| M_{A a} \cap F_{Aa}) P( M_{A a} \cap F_{Aa})\\ &+ P( C_{A a}| M_{A A} \cap F_{Aa}) P(M_{A A} \cap F_{Aa} )\\ =& \frac{2}{4} 2p(1-p)p^2+ \frac{2}{4} (2p(1-p))^2+ \frac{2}{4} p^2 2p(1-p) \\ =& 2p(1-p)p^2+ \frac{2}{4} (2p(1-p))^2 \\ =&2p^2(1-p) \end{aligned} $$
In sum: The first method produced the answer $P\left(C_{A a} | C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c}\right) =\frac{2 p^{2}(1-p)}{p^{2}(3-2 p)}$ while the second method yielded $P\left(C_{A a} | C_{a a}^{c} \cap M_{a a}^{c} \cap F_{a a}^{c}\right) =2 p^{2}(1-p)$. The second method captured only the numerator of the first method. What am I missing?