Mean and Variance of Number of Heads

Question

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?

Answer 1

In Law of Total Variance, $Y$ is a random variable. So for simplicity the definition should be like

$$ Y = \begin{cases} 1 & \text{if Coin 1 is chosen} \\ 0 & \text{if Coin 2 is chosen} \end{cases}$$

And thus $$ Var[X \mid Y] = \begin{cases} \displaystyle n \times \frac {1} {3} \times \frac {2} {3} = \frac {2n} {9} & \text{if} & Y = 1 \\ \displaystyle n \times \frac {1} {4} \times \frac {3} {4} = \frac {3n} {16} & \text{if} & Y = 0 \end{cases}$$

You may write this compactly as

$$ Var[X \mid Y] = \frac {2n} {9} Y + \frac {3n} {16}(1 - Y)$$ The expectation will be $$ E[Var[X \mid Y]] = \frac {2n} {9} \times \frac {1} {2} + \frac {3n} {16} \times \frac {1} {2} = \frac {59n} {288} $$

as the question presumed assumed both coins are equally-likely to be chosen.

Similarly, $$ E[X \mid Y] = \frac {n} {3} Y + \frac {n} {4} (1 - Y) = \frac {n} {4} - \frac {n} {12} Y$$ $$ Var[E[X \mid Y]] = \frac {n^2} {12^2} \times \frac {1} {4} = \frac {n^2} {576} $$

Therefore,

$$ Var[X] = \frac {59n} {288} + \frac {n^2} {576} $$