I have the exact same question as the person who asked this question: Determining the probability mass function of an internet service provider using 50 modems to serve 1000 customers.
but I couldn't comment on the provided answer (low reputation). But I don't think the one single answer to the post answers why we need to sum over all the possibilities, if there are $\geq$ 50 customers who needs a connection we need to sum from $50, ..., 1000$: $$ P(X = 50) = \sum_{k=50}^{1000} {1000 \choose k} 0.01^k (0.99)^{1000 - k} $$
Could somebody elaborate more on as to why we need to do the summation?
Because for when $k < 50$, this is not the case: $$ P(X = k) = {1000 \choose k} 0.01^k (0.99)^{1000 - k}, \hspace{2 cm} k = 0..49 $$
which I don't fully understand why.
The problem is ambiguously stated. The assumption there is "customer in need of a connection is assigned to an unused device if available". With that clarification the first customer uses any 1 modem, second any other modem (and two are used) and with $k\leq 49$ it's a matter of k successes in a Binomial distrubution with probability p and 1000 trials.
With 50 or more customers there are no unused modems, X=50 for 50 successes (50 customers), 51 successes (51 customers) and so on, the probabilities need to be summed. We can also say $P(X=50)=1-\sum_{c=0}^{49}P(c\ customers\ need\ a\ connection)$, but with these values the sum will round up to 1 with low precision.
If customers could be assigned to the same modem with unused modems available, the problem is more interesting.