# Is this a negative binomial?

• March 26th 2010, 08:06 PM
mehn
Is this a negative binomial?
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.
(Nerd)
• March 26th 2010, 08:16 PM
Debsta
Quote:

Originally Posted by mehn
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.
(Nerd)

It is not a negative binomial distribution because the trials are not independent (ie the outcome of one trial affects the next).
• March 26th 2010, 08:23 PM
mehn
If not negative binomial distribution, then which distribution should i use?
• March 26th 2010, 08:34 PM
Debsta
Quote:

Originally Posted by mehn
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.
(Nerd)

Quote:

Originally Posted by mehn
If not negative binomial distribution, then which distribution should i use?

I'd just use the multiplication principle.
• March 26th 2010, 08:42 PM
mehn
multiplication principle?
no distribution comes in play here?
• March 26th 2010, 08:49 PM
Debsta
Quote:

Originally Posted by mehn
multiplication principle?
no distribution comes in play here?

Not sure to be honest - not an expert on this stuff - but love probability!

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
• March 26th 2010, 08:53 PM
mehn
Actually this seems to be ok... But i prefer to use some distributions...
Thank you for your help btw =]

Can someone else help me on this?
• March 26th 2010, 09:07 PM
matheagle
I would view this as hypergeometric and another selection.
You need to choose 5 good and 3 bad players on the first 8 selections.
That leaves you with 6 good and 1 bad player.
Hence.......

$\left({{11\choose 5}{4\choose 3}\over {15\choose 8}}\right)\left({1\over 7}\right)$

$=\left({(462)(4)\over 6435}\right)\left({1\over 7}\right)$

$\approx .041025641$