It is expected that a computer will fail after it is on for 8 hours.
(i)Determine the probability that the computer will have at least one failure in a 24 hour day.
(ii) A lab contains 20 such computers. Determine the probability that at least one computer will fail in that lab in one day.
(iii) Determine the probability that between 4 and 16 computers will fail during a particular day.
-for the first part of the question , I used poisson distribution formula
8 hours ---> 1 time broken 24 hours ---> 3 times broken
let X be the number of times the computer fails P(at least one ) = P(X>= 1) = 1-P(X<1)
using mean as 3 for this question 1-P(X<1) = 1 - ( e^-3)(3^0/0!) = 0.9502
-for the second part , when using approximation to binomial , n = 20 p = 0.9502 I am getting the probability = 1 , which seems to be incorrect
For a single computer, the expected failures in a day are $\mu=3$. The Poisson probability that there is at least one failure is $P(k\geq 1; \mu)= 1-P(k=0; \mu) = 1 - e^{-\mu} \approx 0.95$ as you find.
The previous result, for a single computer, is quite large, so if there are $n=20$ computers, we can expect the chance that at least one fails to be even closer to one. I would write $P(k\geq 1; \mu, n) = 1- P(k=0; \mu, n) = 1-(P(k=0; \mu))^n = 1 - e^{-n \mu} \approx 1$.