The agony of group gifts

Because there are $n$ independant calls with probability $p$ of success, $X$ follows a binomial distribution $\mathcal{B}(n,p)$:
$$P(X=i)=\binom{n}{i}p^{i}(1-p)^{n-i}$$

In the second round, the calls are still independant and each with probability $p$ to succeed, but this time, if $X=i$, there are only $n-i$ calls. So the conditional probability is again a binomial distribution $\mathcal{B}(n-i,p)$:
$$P(Y=k\mid X=i)=\binom{n-i}{k}p^{k}(1-p)^{n-i-k}$$

Disclaimer : there is probably a better way than the brute force method I did, but I can’t find the right interpretation: when $X=i$, then we have to have $Y=k-i$. This means that :
$$P(X+Y=k)=\sum_{i=0}^{k}P(X=i\quad\text{and}\quad Y=k-i)$$But these joint probabilities can be expressed using the conditional probabilities which were computed in the previous question, and using the first question:
$$P(X+Y=k)=\sum_{i=0}^{k}P(X=i)P(Y=k-i\mid X=i)=\sum_{i=0}^{k}\binom{n}{i}p^{i}(1-p)^{n-i}\binom{n-i}{k-i}p^{k-i}(1-p)^{n-k}$$and a few terms do not depend on $i$, so they can be extracted from the sum:
$$P(X+Y=k)=p^{k}(1-p)^{n-k}\sum_{i=0}^{k}\binom{n}{i}\binom{n-i}{k-i}(1-p)^{n-i}$$Here there is a nice combinatorial identity:
$$\binom{n}{i}\binom{n-i}{k-i}=\binom{k}{i}\binom{n}{k}$$which yields:
$$P(X+Y=k)=p^{k}(1-p)^{n-k}\binom{n}{k}\sum_{i=0}^{k}\binom{k}{i}(1-p)^{n-i}$$This sum is very close to a binomial type of expression. We make a $k$ appear:
$$P(X+Y=k)=p^{k}(1-p)^{n-k}\binom{n}{k}\sum_{i=0}^{k}\binom{k}{i}(1-p)^{n-k}(1-p)^{k-i}=p^{k}\left((1-p)^{n-k}\right)^{2}\binom{n}{k}\sum_{i=0}^{k}\binom{k}{i}(1-p)^{k-i}$$$$=p^{k}\left((1-p)^{n-k}\right)^{2}\binom{n}{k}(1-p+1)^{k}=p^{k}\left((1-p)^{n-k}\right)^{2}\binom{n}{k}(2-p)^{k}$$By letting $q=p(2-p)=2p-p^{2}$, we have that $1-q=1-2p+p^{2}=(1-p)^{2}$, we just obtained:
$$P(X+Y=k)=\binom{n}{k}q^{k}(1-q)^{n-k}$$This is again a binomial distribution $\mathcal{B}(n,q)$.

The expected value of $Z=X+Y$ is $nq$, while the variance is $nq(1-q)$.