# 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?

UPDATE: I left out some information in the problem the first time this was posted, and I apologize.