A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one, and examined. The ones examined are not put back. What is the probability that the 9th one examined is the last defective?
I have tried doing this by using the negative binomial but i'm not reaching the answer. I can't find any other distribution for doing this.
Any help please?
I just looked at the probability of getting 5 good ones and then 4 faulty ones. This gave me:
11/15 * 10/14 * 9/13 * 8/12 * 7/11 * 4/10 * 3/9 * 2/8 * 1/7
But this was ordered.
The last one (ninth one) had to be faulty. So I then looked at how many ways there were to arrange the first 8; 5 good ones and 3 faulty ones ie 8C3.
Then multiplied the result above by that....got 0.04 approx