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Pavel314

Joined: 29 Apr 2005
Posts: 78

Posted: Sun Jun 11, 2006 1:16 pm    Post subject: Re: Defects Problem

"Jonas" <sundet@yahoo.com> wrote in message
 Quote: In order to get a deeper understanding for the problems that I have been doing, I try to look at them in a couple of different ways. One of the problems is the following: A box contains 24 light bulbs of which 4 are deffective; 1/6 of the light bulbs are defective. If you choose 4 light bulbs from the box without replacing them what is the probability that all four will be deffective? I know that there are 24 choose 4 combinations of choosing light bulbs. This number would be the denominator. For the numerator, I think that 4 choose 4 should be used. The result from this path is 1/10,626. If I use the binomial formula with p=1/6 and q=5/6, I the answer that I get is (1/6)^4=1/1,296 I'm pretty certain that I am making a fundamental error but I can't figure it out. Your suggestions are appreciated.

You need to use the hypergeometric distribution for selection without
replacement. Details can be found at:
http://en.wikipedia.org/wiki/Hypergeometric_distribution

The basic formula is C(D,k)*C(N-D, n-k) / C(N,n) where N=total objects (24),
D=defective (4), n=sample size (4) and k=defects in sample (4). For your
problem,

C(D,k)*C(N-D, n-k) / C(N,n) =

C(4,4) * C(24-4,4-4) / C(24,4) =

C(4,4) * C(20,0) / C(24,4) =

1 * 1 / 10,626 =

1 / 10,626 =

..0000941 or 0.00941%

A slim chance indeed.

I ran the numbers for other results on a draw of 4 and got:

4 defects = 0.01%
3 defects = 0.75%
2 defects = 10.73%
1 defect = 42.91%
0 defects = 45.60%

These total 100%.

Paul
danheyman@yahoo.com
science forum beginner

Joined: 18 Jul 2005
Posts: 33

Posted: Sat Jun 10, 2006 10:56 pm    Post subject: Re: Defects Problem

The way you did the problem by counting is correct. The binomial
distribution is based
on sampling with replacement, so it overestimates the probability of
getting 4 defectives when sampling without replacement.
Dan Heyman

Jonas wrote:
 Quote: In order to get a deeper understanding for the problems that I have been doing, I try to look at them in a couple of different ways. One of the problems is the following: A box contains 24 light bulbs of which 4 are deffective; 1/6 of the light bulbs are defective. If you choose 4 light bulbs from the box without replacing them what is the probability that all four will be deffective? I know that there are 24 choose 4 combinations of choosing light bulbs. This number would be the denominator. For the numerator, I think that 4 choose 4 should be used. The result from this path is 1/10,626. If I use the binomial formula with p=1/6 and q=5/6, I the answer that I get is (1/6)^4=1/1,296 I'm pretty certain that I am making a fundamental error but I can't figure it out. Your suggestions are appreciated.
Jonas

Joined: 08 Sep 2005
Posts: 77

 Posted: Sat Jun 10, 2006 4:01 pm    Post subject: Defects Problem In order to get a deeper understanding for the problems that I have been doing, I try to look at them in a couple of different ways. One of the problems is the following: A box contains 24 light bulbs of which 4 are deffective; 1/6 of the light bulbs are defective. If you choose 4 light bulbs from the box without replacing them what is the probability that all four will be deffective? I know that there are 24 choose 4 combinations of choosing light bulbs. This number would be the denominator. For the numerator, I think that 4 choose 4 should be used. The result from this path is 1/10,626. If I use the binomial formula with p=1/6 and q=5/6, I the answer that I get is (1/6)^4=1/1,296 I'm pretty certain that I am making a fundamental error but I can't figure it out. Your suggestions are appreciated.

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