Poisson distribution vs multiplying probabilities I am a TA for a stats course for engineers, and I had a really good question from a student today, which I don't know the answer to.
We were going through the following word problem:
"4 computers run continuously for the Toronto Stock Exchange. The probability of a computer failure in a day is estimated at 5%. Assuming differing computers fail independently, what is the probability that all 4 computers fail in a day?"
Since the sampling takes place over an interval, the way I would approach this is using the Poisson distribution, with the average number of computers failing on a day $\equiv\lambda = 0.05$. If four computers fail, then $k = 4$. Thus,
\begin{align*}
   P(k; \lambda) &= \frac{\lambda^{k} e^{-\lambda}}{k!} \\
   P(k=4; \lambda = 0.05) &= \frac{0.05^{4} e^{-0.05}}{4!} \\
   & = 2.477\times 10^{-7}
\end{align*}
However, a student asked why it would not be appropriate to just multiply the probability of each computer failing. Since the probability of each computer failing each day $\equiv p = 0.05$, and since each computer failure is independent, he argued that,
\begin{align*}
   P(k=4) &= p^4 \\
          &= 0.05^4 = 6.25\times 10^{-6}
\end{align*}
Which one of these approaches is wrong given the question? And why? What underlying assumption of the wrong approach is violated by the question?
Thank you for your help.
UPDATE: I left out some information in the problem the first time this was posted, and I apologize.
 A: The Poisson process that you're using assumes that 0.05 is the expected number of computers failing in one day in an unknown number of total computers (your answer also assumes that this rate is fixed after a computer fails, which implies that computers can fail multiple times, or are replaced immediately, or there are so many of them that this is negligible).
The independent probability that the student is using assumes that there are exactly four computers each of which has a 5% chance of failing.
The wording makes it sound to me like 5% is the chance of any individual computer failing (so the second interpretation).  In that case, we want to know the total number of computers and apply a binomial distribution.  Since the question doesn't give the total number of computers, it can't be answered.
Another possibility is that 5% is the probability that exactly one computer fails, and yet another possibility is that 5% is the probability that at least one computer fails.  In either case you can deduce the Poisson process intensity that gives this value.  For the first of these, I get 4.4997552907483822; for the second, I get an intensity of 0.051293294149203306.  From there you could calculate similarly to how you did.

Per your update:  You can eliminate the Poisson process since you don't have a fixed rate.  You still have to decide whether 5% is the probability of a given computer failing, in which case the student is right.  If it's the probability of at least one computer failing, or the probability of exactly one computer failing, you'll have to reason back from that number to the probability of any individual computer failing before reasoning forwards.
