Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 $4$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.