variance of the estimation $p^*=(x_1*x_2)/(n*n)$

Question

I have that $X_1$~$Bin(n,p_1)$ and $X_2$~$Bin(n,p_2)$ and want to calculate the variance of the estimation $p^*=(x_1*x_2)/(n*n)$, which is an observation of the estimator $p^*(X_1,X_2)=(X_1*X_2)/(n*n)$.

my attempt:

$V(p^*)=V((X_1*X_2)/(n*n))=E(((X_1X_2)/(n^2) - p_1p_2)^2)=(1/n^4)E((X_1X_2)^2) - p^2$

there $p=p_1p_2$ and $E((X_1X_2)^2)=E(X_1^2X_2^2)=E(X_1^2)E(X_2^2)$ since $X_1$ and $X_2$ are independent. Therefore $V(p^*)=(1/n^4)E(X_1^2)E(X_2^2)-p^2$

Here it takes stop for me.

EDIT:

The estimation will be use to estimate the probability of a reliability systems consisting of two subsystems connected in parallel , which breaks independent of each other and have the probabilities p1 and p2 to break in a week. $x_1$ stands for that subsystem 1 has broken $x_1$ times and $x_2$ that subsystem 2 has broken $x_2$ times

Answer 1

You're almost there, to conclude, you just have to notice that

$$E(X_1^2)=V(X_1)+E(X_1)^2=np_1(1-p_1)+(np_1)^2=np_1(1+(n-1)p_1)$$

and similarly $$E(X_1^2)=np_2(1+(n-1)p_2)$$

therefore

$$V(p^*)=\frac{1}{n^2}p(1+(n-1)p_1)(1+(n-1)p_2)-p^2.$$