We know that there are 15 bad brakes in a lot of 500. We need to pick 4 brakes for use.
The probability of picking 1 bad brake = 15/500 = 3%
The probability of picking 4 bad brakes = 15/500+14/499+13/498+12/497 = 10.83 %

Am I right ?

15/500 *14/499*13/498*12/497= very unlikely the 4 u picked are all bad

The probability of picking a single bad brake is indeed

15/500 = 3%.


However, the probability of picking a bad brake followed by 3 good brakes is

15/500 * 485/499 * 484/498 * 483/497 = 2.75%

But the bad one could be drawn 1st, 2nd,3rd or 4th, so multiply by 4 -> 11.0% that one of the four is bad!


You can come mighty close just using the binomial distribution. In Excel that would be =BINOMDIST(1,4,0.03,FALSE) which gives 10.9% Binomial assumes that there is a 3% chance that the next is bad, no matter what the ones before were. Since the lot is large (500) and the sample is small (4), this approximation should worke pretty well.


I agree with roland_lu for the odds of all four being bad.
Tim F

If you want to come somewhat closer than the binomial, use the hyper geometric. The hypergeometric does not cumulate in Excel, so you have to write out the probabilities for 0, 1, 2, 3, and 4 bad brakes. Or simply subtract the probability of no bad brakes from 1.

Please see attached Excel file.