Finding the Probability of a Probability Using binomial

Question

The exercise goes like this:

Let $X$ be $Bin(n,p)$ where $n=10$ and $p$ can be either $1/4$ or $3/4$, where both possibilities of $p$ are equally likely. It is known that $X=7$. What is the probability that $p=3/4$

If $P(X=7)= \binom{10}{7} p^7 (1-p)^3$ Do I have to solve some sort of inequality like: $\binom{10}{7} (1/4)^7 (3/4)^3 = \binom{10}{7} (3/4)^7 (1/4)^3 =1/2$ ?

Answer 1

You can do this with Bayes' theorem: $$P(A|B)=P(B|A)\frac{P(A)}{P(B)}$$ where $A$ is the event $p=\frac 34$, and $B$ is the event $X=7$. All of $P(A|B)$, $P(A)$ and $P(B)$ can be calculated, so you can calculate $P(A|B)$ too.

Note that $$P(B)=P(X=7)=P\left(X=7\left|p=\frac 14\right.\right)P\left(p=\frac 14\right) +P\left(X=7\left|p=\frac 34\right.\right)P\left(p=\frac 34\right) $$