# 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 their only kid blood phenotype is A.

Then what is the probability $$\mathbb{P}(X|Y)$$?

Here is my analysis:

From Bayess 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.

Thank you for precious time on this matter.

Reference

• Something worth keeping in mind is that the height of a valid probability density function (PDF) can exceed $1$. What can’t exceed $1$ is the integral of the PDF.
– Dave
Aug 12 at 7:16

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 $$P(X)$$.