Prove the sum of 2 Bernoulli is binomial let $X_1,X_2$ be independent and identically distributed Bernoulli random variables with parameter p ∈ (0, 1).Prove that the sum $Y=X_1+X_2$ of these random variables is a binomial random variable with parameters 2 and p
 A: If $X_1, X_2$ are two iid Bernoulli random variables, then their sum is the count of successes in two iid Bernoulli experiments.   This is the definition of a binomial random variable  with parameters $2$ and $p$.
That is all. $\Box$

If you need to demonstrate further:
The probability that there will be exactly $y$ successes among these two variables is determined by measuring the probability of $y$ successes in a row followed by $2-y$ failures, then multiply this by the count of distinct ways to order these results.
That is:
$$\mathsf P(Y=y) = \underline{\qquad\qquad?}$$
Which was to be demonstrated. $\Box$

Alternatively, from first principles we use the Law of Total Probability to show that when we partition the results by the first variable:
$$\begin{align}
\mathsf P(Y=y)
 &= \mathsf P(X_1=1, X_2=y-1) + \mathsf P(X_1=0, X_2=y)
\\[1ex]& = p\cdot \mathsf P(X_2=y-1) + (1-p)\cdot \mathsf P(X_2=y)
\\[1ex]& = \begin{cases}
 \underline\qquad & : y =0
\\ \underline\qquad & : y=1
\\ \underline\qquad & : y=2
\end{cases}
\end{align}$$
Which was to be demonstrated. $\Box$

Fill in the blanks if required.  But really, you don't need to go passed the first tombstone.
A: The range of the random variable must be $\{0,1,2\}$.  You have to show the probabilities of the three values is the same.  So just calculate the three probabilities of a binomial with parameters $2$ and $p$ and do the same for the sum of two Bernoulli trials with parameter $p$ and you'll see the three probabilities are the same in both cases.  So they are the same random variable.
