Finding the probability that the sum of two binomial random variables is $1$. 
Let $X_1$ and $X_2$ be independent binomial random variables with $X_i$ having parameters $\left(n_i,p_i\right),\,i=1,2$. Find $\mathrm{Pr}\left(X_1+X_2=1\right)$.

$$\begin{aligned}
P\left(X_1+X_2=1\right)&=P\left[\left(X_1=1\cap X_2=0\right)\cup\left(X_1=0\cap X_2=1\right)\right]\\&= P\left(X_1=1\cap X_2=0\right)+P\left(X_1=0\cap X_2=1\right)-\\&-P\left(X_1=1\cap X_2=0\cap X_1=0\cap X_2=1\right)\\
&=P(X_1=1)P(X_2=0)+P(X_1=0)P(X_2=1)-\\
&-P(X_1=1)P(X_2=0)P(X_1=0)P(X_2=1).
\end{aligned}$$
Is this correct so far?

That's almost true, except for the very last term. The event $X_1=1$ is not independent from $X_1=0$, to put it mildly, so you can't write $P(X_1=1\cap X_1=0)=^?P(X_1=1)P(X_1=0)$.
