Can a posterior probability be larger than 1 when more samples are available? A "genotypes - phenotypes blood" question I am working on is
In a family, Both father's and mother's blood phenotype is A.
Assume

*

*the event X is both father's and mother's blood genotype is AA.

*the event Y is only one of their kids blood phenotype is A.

Then what is the probability $\mathbb{P}(X|Y)$?
Here is my analysis:
From Bayes's theorem, $\mathbb{P}(X|Y) = \mathbb{P}(\frac{\mathbb{P}(Y|X)\mathbb{P}(X)}{\mathbb{P}(Y)})$
1.
given the blood phenotype is A,
there are four combinations, (AA, AA), (AA, AO), (AO, AA), (AO, AO)
So
$\mathbb{P}(X)$ is $\frac{1}{4}$
2.
Because both father's and mother's genotypes blood are then
$\mathbb{P}(Y|X)$ is $1$
3.
$\mathbb{P}(Y)$ = $\mathbb{P}(X)\mathbb{P}(Y|X) + \mathbb{P}(\bar{X})\mathbb{P}(Y|\bar{X})$
Where $\mathbb{P}(\bar{X})$ is $\frac{3}{4}$
$\mathbb{P}(Y|\bar{X})$ is $\frac{11}{12}$
(in (AA, AO), (AO, AA), (AO, AO), there are 12 cases and only one case is O)
Then $\mathbb{P}(Y)$ is $\frac{1}{4}$ + $\frac{3}{4}*\frac{11}{12}$ = $\frac{15}{16}$
4.
So $\mathbb{P}(X|Y)$ is $\frac{\frac{1}{4}}{\frac{15}{16}}$ is about 0.26666, which is larger than $\mathbb{P}(X)$ = 0.25
However,
if they have more kids, like 26 kids, and assume each kid's blood phenotype is independent of each other,
$\mathbb{P}(X|Y)$ = $\frac{\frac{1}{4}1^{26}}{(\frac{15}{16})^{26}}$ is about 1.3387, which is larger than 1.
Can you please help me point out which step is incorrect?
Thank you for your precious time on this matter.
Reference

*

*What are genotypes and phenotypes for blood

2022.12.28 Update
With @frank 's very important suggestions,
I figure out my mistake, and here is the correction (please let me know if there is still any mistake)
Assume

*

*the event X is both father's and mother's blood genotype is AA.


*the event Y is only one of their kids' blood phenotype is A.


*the event Z is two of their kids' blood phenotype is A.


*each genotype has an equal probability: AA, AO, BB, BO, AB, OO. $\mathbb{P}(genotype=AA)$ = $\mathbb{P}(genotype=OO)$ = $\frac{1}{6}$


*each kid's blood type (phenotype or genotype) is independent from each other
Therefore
1.
$\mathbb{P}(X)$ = $\frac{1}{6}$ * $\frac{1}{6}$ = $\frac{1}{36}$
2.
$\mathbb{P}(Y)$ = $\frac{40}{144}$ = $\frac{5}{18}$
3.
$\mathbb{P}(X|Y) = \mathbb{P}(\frac{\mathbb{P}(Y|X)\mathbb{P}(X)}{\mathbb{P}(Y)})$, where $\mathbb{P}(Y)$ = $\mathbb{P}(X)\mathbb{P}(Y|X) + \mathbb{P}(\bar{X})\mathbb{P}(Y|\bar{X})$
So we have
$\frac{1*\frac{1}{36}}{1*\frac{1}{36}+\frac{36}{140}*\frac{35}{36}}$ = 0.1 = 10% (which is more "certain", comparing to {P}(X)=0.027$^{.}$, about 2.78%)
4.
from Bayes Theorem with multiple conditions
and the independence assumption,
$\mathbb{P}(Z) = \mathbb{P}(X|Y_1 \cap Y_2)$
= $\mathbb{P}(\frac{\mathbb{P}(Y_1 \cap Y_2|X)\mathbb{P}(X)}{\mathbb{P}(Y_1 \cap Y_2)})$,
where the denominator is
$\mathbb{P}(Y_1 \cap Y_2)$ = $\mathbb{P}(Y_1 \cap Y_2|X)$$\mathbb{P}(X)$ + $\mathbb{P}(Y_1 \cap Y_2|\bar{X})$$\mathbb{P}(\bar{X})$
= $\mathbb{P}(Y_1|X)$$\mathbb{P}(Y_2|X)$$\mathbb{P}(X)$ + $\mathbb{P}(Y_1|\bar{X})$$\mathbb{P}(Y_2|\bar{X})$$\mathbb{P}(\bar{X})$
= $\frac{1}{36}$ + $(\frac{36}{140})^{2}$ * $\frac{35}{36}$
and the numerator is
$1*\frac{1}{36}$
$\mathbb{P}(Z) = 0.30172, about 30.17%
5.
With "more kids' phenotype is A" evidence,
n in $(\frac{36}{140})^{n}$ goes infinite,
and the denominator converges to $\frac{1}{36}$
Therefore, it satisfies that

*

*with more evidence (observations), and more certainty.

*the probability will not exceed 1.

Again, thank you for your precious time on this long discussion.
Thank you, @frank.

As a rule, the probability of an event is never greater than one (the pdf can take any arbitrarily high value (nonnegative), but the probability measure of any event, i.e. any member of the $\sigma$-algebra, cannot be larger than one).
It makes sense to compute $P(X | Y)$ with the Bayesian theorem, but then you need the probabilities $P(X)$ and $P(Y)$, which you don't know. You don't know the probability that both parents have AA, nor that a child has phenotype A.
Your attempt to e.g. compute $P(X)$ in point (1) could be improved: You have presumed all those pairings to have the same probability (which they don't have) and you didn't consider the fact that (AO, AO) can also lead to the child being O. And even then you would have computed $P(X|Y)$ not $P(X)$.
There are some additional issues in your post, e.g. the argument with the 26 kids and the computation of the 26th power of a number less than one giving a number larger than one.
