I am considering the following setting:
I have $n$identical boxes. I have $m$people come sequentially. Each people will randomly choose one box with probabiliy $p$ and choose nothing with probability $1-p$. Now I am interested in the probability that a box has $i$ people chosen by the end of the game, i.e., all $m$ people have made their decisions.
I was wondering if this probability is
\begin{align*} &P\{box~1~has~i~people\}\\ &=\sum_{j=i}^{m}P\{box~1~has~i~people|j~people~choose~some~box\}P\{j~people~choose~some~box\}\\ &=\sum_{j=i}^{m}\binom{j}{i}(1/n)^i\binom{m}{j}p^j \end{align*} Is this derivation correct?
No, that's not quite correct. You haven't accounted for the fact that the other $m-i$ people must not choose the box.
There's no need to do this in two steps with a summation. Each person has probability $\frac pn$ to choose the box, so the probability for exactly $i$ people to choose it is simply
$$ \binom mi\left(\frac pn\right)^i\left(1-\frac pn\right)^{m-i}\;. $$