Determining the probability mass function of an internet service provider using 50 modems to serve 1000 customers. Hello I am attempting to understand the solution to the following problem.
Given: An internet service provider uses 50 modems to serve the needs of 1,000 customers. It is estimated that at a given time, each customer will need a connection with p = 0.01, independent of the other customers.
Question: What is the probability mass function of the number of modems in use at the given time?
Solution:
Let $X$ be the number of modems in use. For $k < 50$, the probability that $X=k$ is the same as the probability that $k$ out of $1000$ customers need a connection is: 
$$p_X(k) = \binom{1000}{k}(0.01)^k(0.99)^{1000-k} \quad \textrm{for} \quad  k = 0, 1,..., 49.$$
The probability that $X = 50$ is the same as the probability that $50$ or more out of $1000$ customers need a connection is:
$$ p_X(50) = \sum\limits_{k=50}^{1000}\binom{1000}{k}(0.01)^k(0.99)^{1000-k}$$
My question: When I originally tried solving this problem I wrote $\binom{50}{k}$ instead of $\binom{1000}{k}$ because I thought that we had to choose k modems out of 50 possible that can be in use at a given time. Then I realized that this didn't make sense because $(0.01)^k(0.99)^{1000-k}$ represents the probabilities of $k$ people that need or don't need a connection. However I don't understand why "the probability that $X=k$ is the same as the probability that $k$ out of $1000$ customers need a connection". If I had 49 modems in use e.g., isn't it be possible that said 49 modems provide connections for any number of people up to 999? 
Why does it make sense or why is it intuitive to equate the probability of 49 modems being in use with the probability that 49 people require a connection?
Further, why do we sum the probabilities from $k=50, ...., 1000$ and not for $k=0,1,...,49$?
Thanks for the help!
 A: First of all the sum of all probabilities has to be $1$. This is here the case:
$P(X=k)= \binom{1000}{k}(0.01)^k(0.99)^{1000-k} \quad \textrm{for} \quad  k = 0, 1,..., 49$ 
$P(X= 50)= \sum\limits_{k=50}^{1000}\binom{1000}{k}(0.01)^k(0.99)^{1000-k} $
Also note that $X$ is the random variable for the modems in use. If $k$ (cutomers who need connections) is greater then $50$ there are still only $50$ modems in use.

If I had 49 modems in use e.g., isn't it be possible that said 49
  modems provide connections for any number of people up to 999?

That´s true.

Why does it make sense or why is it intuitive to equate the
  probability of 49 modems being in use with the probability that 49
  people require a connection?

The numbers of modems in use is only equal if the number of people require a connection do not exceed the number of available modems. If you have more people require a connection than available modems then every modem is in use but some/many people don´t get a connection.

Further, why do we sum the probabilities from $k=50, ...., 1000$

We need the probability that more than 50 or equal to 50 users want to get a connection. In this case all 50 modems are in use.

and not for $k=0,1,...,49$?

Because it is a pdf and not a cdf.
