Probability Math
posted by Jack on .
Through a mix up on the production line, 5 defective light bulbs were shipped out with 32 good ones. If 4 are selected at random, what is the probability that at least 3 of them are defective?

This typically calls for the hypergeometric distribution, which gives
P(y)=[(r,y)(Nr,ny)]/(N,n)
where
N=total 32+5=37 light bulbs
n=sample size of 4
r=defective out of N, 5
y=number of defective in the sample
(x,y)=combination=x!/(y!(xy)!)
P(3)=[(5,3)(375,53)]/(37,5)
=4960/435897
P(4)=160/435897
Total P(3+)=5120/435897=1.17%