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Forum index » Science and Technology » Math » Probability
Defects Problem
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Jonas
science forum addict


Joined: 08 Sep 2005
Posts: 77

PostPosted: Sat Jun 10, 2006 4:01 pm    Post subject: Defects Problem Reply with 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.
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danheyman@yahoo.com
science forum beginner


Joined: 18 Jul 2005
Posts: 33

PostPosted: Sat Jun 10, 2006 10:56 pm    Post subject: Re: Defects Problem Reply with quote

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.
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Pavel314
science forum addict


Joined: 29 Apr 2005
Posts: 78

PostPosted: Sun Jun 11, 2006 1:16 pm    Post subject: Re: Defects Problem Reply with quote

"Jonas" <sundet@yahoo.com> wrote in message
news:1149955268.060295.299770@y43g2000cwc.googlegroups.com...
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
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