binomial distribution and function of several random variables Let $X_1$ and $X_2$ be independent random variables with respective binomial distributions $b(3,1/2)$ and $b(5,1/2)$. Find $P(X_1+X_2=7)$.
I am thinking of an example I pulled from my notes:
Ex: let $X_1$ ~ Poisson $(\lambda_1=2)$ and $X_2$~Poisson$(\lambda_2=3)$ be independent, then $P(X_1+X_2=1)=P(X_1=0,X_2=1)+P(X_1=1,X_2=0)=(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)}$
So my thoughts for this question: $P(x,p)=\binom{n}{x}(p)^x (1-p)^{n-x}$. So for this question, should I cconsider by cases, like $X_1=1, X_2=3$ etc, and since they are independent, then I would add those cases together?
 A: The binomial distribution with amount $n$ and success rate $p$, is more often denoted $\mathcal{Bin}(n,p)$.
$Z\sim\mathcal{Bin}(n,p)$ denotes that random variable, $Z$ is a count of successes among $n$ independent Bernoulli trials with identical success rate $p$.
So given that the independent random variables are $X_1\sim\mathcal{Bin}(3, 1/2)$ and $X_2\sim\mathcal{Bin}(5,1/2)$, what does $X_1+X_2$ count?   What is the distribution for this?

$X_1$ counts successes among $3$ independent Bernoulli trials with identical success rate $1/2$, and independently $X_2$ counts successes among $7$ independent Bernoulli trials with identical success rate $1/2$.
So $X_1+X_2$  counts...

Use this.

Okay, it is possible to use the Law of Total Probability.
Now, when the sum is $7$ the minimum $X_1$ may be is $2$ (because the maximum $X_2$ may be is $5$), so …

 $$\begin{align}\mathsf P(X_1+X_2=7)~&=~\sum_{k=2}^3\mathsf P(X_1=k)~\mathsf P(X_2=7-k)\\[1ex]&=~\mathsf P(X_1=2)\,\mathsf P(X_2=5)+\mathsf P(X_1=3)\,\mathsf P(X_2=4)\\[1ex]&~~\vdots\end{align}$$

