Its about 32,000 rounded up
We can use the binomial theorem to find the probability that 0 out of the 15 samples will be defective, given that 20% are defective.
P(0/15) = (15C0) (0.2)^0 (1 - 0.2)^15 = (1)(1)(0.8)^15 = 0.0352
Then the probability that at least 1 is defective is equal to 1 - 0.0352 = 0.9648. This means there is a 96.48% chance that at least 1 of the 15 samples will be found defective. This is probably sufficient, though it depends on her significance level. If the usual 95% is used, then this is enough.
The answer is:
It’s equivalent to 30+5y
