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

Answer 1

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.

Answer 2

A = .0498

B = .5768