Question

F.I. Determine the probability that there are 3 defective items in a sample of 100 items if 2% of items made in this factory are defective...​

Answer 1

p(defective) =0.02

p (not defective) = 0.98

a) (0.98)^47 * (0.02)^3 = 0.38692 * 0,000008 = 3.09 * 10^-6

or 0.000309%

The order does not come into play here, any 3 bulbs can be defective.

b) The chance for no defective bulbs in  a sample of 50:

(0.98)^50, or about 0.36417 or 36.42%