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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 might be able to somehow deduce the Poisson process intensity that gives this value. From there you could calculate similarly to how you did.

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.

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 could deduce the Poisson process intensity that gives this value (using the inverse cdf). From there you could calculate similarly to how you did.

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 might be able to somehow deduce the Poisson process intensity that gives this value. From there you could calculate similarly to how you did.

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 computers failing multiple times is allowed or else there is such a large number of computers 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.

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 could deduce the Poisson process intensity that gives this value (using the inverse cdf). From there you could calculate similarly to how you did.