Author 
Message 
Pavel314 science forum addict
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
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(ND, nk) / 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(ND, nk) / C(N,n) =
C(4,4) * C(244,44) / 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 

Back to top 


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. 


Back to top 


Jonas science forum addict
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. 

Back to top 


Google


Back to top 



The time now is Tue Oct 24, 2017 12:36 am  All times are GMT

