For this, I would use a Poisson distribution. (There are other methods).
Under a Poisson, the probability of observing exactly k true events is:
P=[y^k * e^(-y)]/k!
where y is the expected number = 3
So, P(0) = [3^0 * e^(-3)/0! = .04979
P(1) = 3^1 * e^(-3)/1! = .14946
and so on.
Take it from here.
Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a random sample of 200 antennas, find the probability that 5 or more will be defective
2 answers
A = .0498
B = .5768
B = .5768