Question 1020364
This is an example of a rare event, so the Poisson distribution is the most appropriate to use.  

{{{p(x) = (mu^x/x!)*e^(-mu)}}}

The {{{mu}}} in this case is equal to 4.

==>The probability of 3 or fewer defective coins would be p(0) +p(1) + p(2) + p(3) = {{{((4^0/0!)+(4^1/1!)+(4^2/2!)+(4^3/3!)) *e^(-4) = 0.43347}}}, to 5 decimal places.

==> The probability of more than 3 defective coins is 1 -  0.43347 = 0.56653.

The probability of exactly 4 defectives is {{{p(4) =  (4^4/4!)*e^(-4)=0.19537}}}, to 5 decimal places.

For the Poisson distribution, {{{mu = sigma^2 = 4}}}.