Likelihood of a data point being in one of two binomial distributions

Question

I have two binomial distributions A and B, and a number v. v must belong to either A or B. I would like to calculate the percentage chance v belongs to distribution B.

My initial, rather naive, approach is to look at the relative curve values at that point:

mean = series_a.mean()
std = series_a.std()
a_percent_likelyhood = st.norm.pdf(number, mean, std)

mean = series_b.mean()
std = series_b.std()
b_percent_likelyhood = st.norm.pdf(number, mean, std)

chance_of_b = b_percent_likelyhood / (b_percent_likelyhood + a_percent_likelyhood)

But as v gets further away from the two distributions, that result tends to a 50% likelyhood of belonging to B, whereas my feeling is that it should tend to 100% on the "side" of B's mean, and tend to 0% on the side of A's mean.

Is this approach indeed flawed, and if so what would be a good alternate solution?

Answer 1

I assume you have the same $N$ for both distributions, but this is not that essential.

By Bayes's rule:

$p(a|v) = \frac{p(v|a)p(a)}{p(v|a)p(a) + p(v|b)p(b)}$

If we assume both distribution $a$ and $b$ were equally likely before we saw the data this becomes:

$p(a|v) = \frac{p(v|a)}{p(v|a) + p(v|b)}$

Exactly what $v$ you are referring to is not totally clear, but I'm guessing you would want to plug in the pmf of distribution $a$ evaluated at $v$ for $p(v|a)$.