How to calculate the probability of "x" things happening out of "n"? I am currently studying for my upcoming midterm and I am stumped on this example provided in the slides. Basically here is the question:

Given 35 computers, what is the probability that more than 10 computers are in use(active)? We are told that each computer is only active 10% of the time. The answer given in the slide is .0004

Here is my following attempt to reproduce that answer:
$$1-{35 \choose 10} \cdot (0.10)^{10} \cdot (0.90)^{25}$$ 
First I got the probability of exactly 10 computers being active out of 35 and then I subtracted 1 from it to get the probability of more than 10 computers. 
EDIT: I have solved this now with the following new work!
1 - (summation of (35 choose k)*0.1^k * 0.9^(35-k) from 0 to 10
 A: As you have correctly noted, the probability of exactly 10 being active at any one time is
$${35 \choose 10} \cdot (0.10)^{10} \cdot (0.90)^{25}.$$
More generally, the probability of exactly $k$ being active at any time is
$${35 \choose k} \cdot (0.10)^{k} \cdot (0.90)^{35-k}.$$
Since you want the probability of more than $10$ being active, you want either
$$1-\sum_{k=0}^{10} {35 \choose k} \cdot (0.10)^{k} \cdot (0.90)^{35-k}$$
(that is, one minus the probability of ten or fewer) or
$$\sum_{k=11}^{35} {35 \choose k} \cdot (0.10)^{k} \cdot (0.90)^{35-k}.$$
A: The expression you have subtracted from $1$ is the correct computation of the probability that exactly $10$ computers are active. The question asks for the probability that more than $10$ are active, so you should sum the chances that $11, 12, 13, \dots$ are active. As you only expect $3.5$ to be active, these will decrease rapidly.  I would compute the chance that exactly $11$ are active (like you did for $10$), then $12$, and keep going until the chance was small enough that it didn't matter any more.
A: You have correctly identified this count as having a Binomial Distribution.
So far, so good.   However, what happened next was not okay.
The complement of having more than $10$ computers active is not of having exactly 10 computers active.   It is of having $10$ or less computers active.
$$\mathsf P(X>10) = 1- \mathsf P(X\leq 10) \\ = 1-\sum_{k=0}^{10} \dbinom{35}{k} 0.10^k~0.90^{35-k}$$
That's a sufficient answer for an exam.   It's a little awkward to calculate the answer.

 $$0.999~575~702~404~549~174~279~490~538~848~808~6$$


Alternatively you could use the Normal approximation to Binomial and lookup Z-tables if they are provided.
