Probability question: Poisson Distribution I am having a problem solving a probability question. At the moment I tried using the poisson distribution but I am getting the incorrect result. Can someone please help me solve the question.
The question reads: 
12. ℎ  ℎ     ℎ ℎ     0.5%.120  
   ,ℎ  ℎ  ℎ 4  ℎ  ℎ.
 ℎ     
My solution is:

 A: This problem is poorly worded, in (at least) three ways.  First of all, we are given the probability that a student is caught cheating, but then asked about the number that cheat.  Anyone who has taught a class of $120$ knows that "cheating" and "getting caught" are not the same.  Secondly, it is not clear whether we are asked about "exactly four" cheaters or "at least four".  I will work the "exactly four" case.  To do the other, you need only compute "exactly  $0,1,2,3$ " and subtract from $1$.  Third,it is not made clear that we are meant to assume that each student is an independent event.  That's not realistic...in my experience, cheaters (especially those that get caught) work in groups.  But as we are told nothing about any dependence we are compelled to assume independence lest the problem be unsolvable.
Exact value:  This is a binomial process with probability $p=.005$.  As such the probability that we get exactly $k$ "successes" is $$p_k=\binom {120}k(.005)^k(1-.005)^{120-k}$$  This is easy to compute (with mechanical assistance) and we get $p_4=\fbox {0.002870399}$
Approximate value:  Poisson is a good idea here, but we have to get the mean right.  As we have $120$ students, we expect $.005\times 120=.6$ of them to be caught.  That's the mean we must use.  Thus the approximation is $$Poisson(4,.6)=\frac {.6^4e^{-.6}}{4!}=\fbox {0.002963583}$$
