How do I find the variance to be able to use chebyshevs inequality? The question is:
The Enterprise has 1000 redundant backups on its warp drive containment system. Each
backup is independent from the others and has a probability of making an error in its
check of 1
200. If more than 50 of the 1000 backups make an error then the warp drive will
lose containment and the Enterprise will explode.
a) Use Chebyshev’s inequality to approximate the chance there are fewer than 50
failures.
b) Use Wolfram Alpha to get a better approximation
I know to use chebyshevs inequality you need the E(x) and the variance. Since its a series of bernoulli trials I know that the E(x) = 5. However, I don't know how to go from here or what I would plug into wolfram alpha to get an approximation.
 A: The probability that a particular backup makes an error is $p$, where I take it $p=1/200$. 
There are $n$ independent backups, so the number of errors is a random variable $X$ that has binomial distribution, with $p=1/200$ and $n=1000$. It is a standard result about the binomial that it has mean $np=5$ and variance $np(1-p)=4.975$.
If you do not know this result, let $X_i=1$ if the $i$-th check fails, and let $X_i=0$ otherwise. We have $X=X_1+\cdots+X_{1000}$. The $X_i$ are independent, so the variance of $X$ is the sum of the variances of the $X_i$. These variances are easy to calculate. 
We are asked for the probability that $X\le 49$.  We have
$$\Pr(X\le 49)=\sum_{k=0}^{49} \binom{1000}{k}(1/200)^k (199/200)^{1000-k}.$$
Wolfram Alpha should be able to compute this without trouble. 
Remark: The question needs some clarification. The preamble says that the spaceship will blow up if there are more than $50$ failures.  
But the actual question asks for the probability of fewer than $50$ failures. The expression we wrote down calculates that. But if we actually want the probability of no more than $50$, the sum would have to be taken to $k=50$. This makes little practical difference, but you should check what the exact wording is. 
