The problem: $X$ is a random variable with $P(X = 1/2) = 1/6$. Given $X = x$, Y is a random variable with binomial distribution with parameters $n = 3$ and $p = x$. My attempt is to find the distribution of $Y$ given $X = 1/2$.
I use the formula by letting $y$ varies:
$P(Y = y | X = 1/2) = \frac{P(X = 1/2|Y = y) P(Y = y)}{P(X = 1/2)}$
I tried with $y = 1$, I got:
$$ P(Y = 1| X = 1/2) = \frac{P(X = 1/2, Y=1) }{P(X = 1/2)} = \frac{\binom{3}{1}(1/2)^{3}}{1/6} = 9/4 $$
Is there anything wrong with my attempt to do the problem here?
You are going wrong if you state that:$$P\left(X=\frac12,Y=1\right)=\binom31\left(\frac12\right)^3$$This should be:$$P\left(Y=1\mid X=\frac12\right)=\binom31\left(\frac12\right)^3$$More general you can state that for $k\in\{0,1,2,3\}$:$$P\left(Y=k\mid X=\frac12\right)=\binom3{k}\left(\frac12\right)^3$$which is exactly what you are looking for: the distribution of $Y$ under condition that $X=\frac12$.