There are 2 coins in a box: the first has probability $\frac 13$of showing heads, the second has probability $\frac14$of showing heads. You choose one coin and you flip it $n$ times (you don’t know which coin you picked). Let $X$ be the number of heads shown by the coin you chose over $n$ flips. Find the mean and variance of $X$.
I found that $E(X)=\frac{7n}{24}$. To find $Var(X)$, I tried using the Law of Total Variance. I defined $Y$to be the event that coin 1 is chosen. Then $$Var(X)=E[Var(X|Y)]+Var[E(X|Y)]=E[n*\frac 13*\frac 23]+Var[n*\frac 13]=\frac 29$$ because since $n$ is fixed, the variance is zero. However this was incorrect. Where did I go wrong?