Posted by **Jack** on Wednesday, May 2, 2012 at 7:16pm.

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?

- Probability Math -
**MathMate**, Wednesday, May 2, 2012 at 11:07pm
This typically calls for the hypergeometric distribution, which gives

P(y)=[(r,y)(N-r,n-y)]/(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!(x-y)!)

P(3)=[(5,3)(37-5,5-3)]/(37,5)

=4960/435897

P(4)=160/435897

Total P(3+)=5120/435897=1.17%

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