# Probability a virus damages 800-850 files of 2400 files with an independent probability of 35%?

A certain computer virus can damage any file with probability 35%, independently of other files. Suppose this virus enters a folder containing 2400 files. Compute the probability that between 800 and 850 files get damaged.

I'm not interested in so much the exact answer as much as I am into how to solve this type of problem. We are working with Poisson distribution and standard deviation which I haven't even figured out how I can even apply to this problem, or if it's even the correct path towards the solution.

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You can model this with a binomial distribution: Let $V$ be the number of files damaged by the virus, we then have:

$$V\sim B(2400,0.35)$$

We then want to compute:

$$P(800\lt V \lt 850)$$

Which can be found using the probability mass function of $V$:

$$P(800\lt V \lt 850)=\sum_{x=801}^{849}{{2400 \choose x}(0.35)^{x}(0.65)^{2400-x}}$$

Which, using Mathematica or a sufficiently large cumulative binomial table (if one even exists!) can be evaluated to $0.613409$, i.e. approximately a $61.3\%$ chance.

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@JeremyQuick: It is likely that you are expected to approximate the probability by finding the probability that a normal with mean $(2400)(0.35)$ and variance $(2400)(0.35)(0.65)$ lies between $800$ and $850$, probably inclusive. –  André Nicolas Apr 6 '13 at 22:48