# Probability of fewer than 5 computers crashing during a thunderstorm?

On average, one computer in 800 crashes during a severe thunderstorm. A certain company had 3000 working computers when the area was hit by a severe thunderstorm.

Compute the probability that fewer than 5 computers crashed?

I'm looking for the steps to solve this problem more than an actual answer so I can work through this on my own. I'm thinking I can solve this using binomial distribution or Poisson distribution but I'm not positive how to go about this for certain.

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Just a note on realism: during a thunderstorm, the events that a certain computer crashes and that a different computer nearby crashes are realistically not independent. So easy-to-work-with distributions are not really what you would work with. But if this is just an exercise... – alex.jordan Nov 20 '12 at 22:27
If you have Excel you can check your answers with the binomdist function – injuryprone Apr 12 '13 at 5:37

Let $p$ be the probability an individual computer crashes, and make the dubious assumption of independence.
Then a reasonable model is that the number $X$ of crashes has binomial distribution, $n=3000$, $p=1/800$. It is relatively straightforward to find $\Pr(X\le 4)$. Some largish numbers are involved, but a calculator that can handle "scientific" notation should have little trouble. Calculate separately $\Pr(X=0)$, $\Pr(X=1)$, and so on up to $\Pr(X=4)$, and add up.
With some thinking, you can cut down on the number of calculator key presses by seeing how to adjust $\Pr(X=i)$ to compute $\Pr(X=i+1$. For calculating up to $4$, however, it may not be worthwhile to search for shortcuts.
Somewhat more pleasant is to notice that $n$ is large, $p$ is small, but $np=3000/800$ is of modest size. So our probability is well approximated by $\Pr(W\le 4)$, where $W$ has Poisson distribution with parameter $\lambda=\frac{3000}{800}$.
You can save key presses by recycling results. For example, $\Pr(X=0)=e^{-\lambda}$. For $X=1$, multiply by $\lambda$. For $X=2$, multiply the $X=1$ result by $\lambda$, divide by $2$. For $X=3$, multiply the $X=2$ result by $\lambda$, divide by $3$. Then comes $4$.