# Hypergeometric distribuion or binomial distribution?

• Feb 1st 2011, 08:00 AM
Mppl
Hypergeometric distribuion or binomial distribution?
There are a set of 500 packages, from this set we know 50 have a defect. In order to inspect the set of packages we select 10 from the 500 packages, testing one by one and putting back the previously tested package before testing another one in order to have always 500 packages to choose, it means, we can test the same package twice or three times, four times,..., ten times.
We reject the set of packages if it happens to test more than 2 packages with defect in our test of 10 packages.

What is the probability of rejecting the set of packages?

I got the right answer using a hypergeometric distribuction, however if I try to use a binomial distribution I don't get the same answer why is that so? with the binomial distribution my procedure was that:

we don't reject the set if we find only zero one or two packages with defect, since we have 50 defective packages in 500 0.1 is the probability of picking up a defective one and 0.9 the probability of picking up a non-defective one then the probability of picking up 0 defective packages is 0.9^10
the prob. of picking up one with defective and 9 nondefective is 0.1*0.9^9 but the one with defect can be the frist the second or the third right? so I have 10 ways of doing that so the probability of having one defective is 10*0.1*0.9^9
and the probability of having two defective is Cn(10,2)*0.1^2*0.9^8
by Cn(10,2) I mean combinations.
suming all the three I have 1-P where P is the prob of rejecting.
the thing is the answer I get back is now right! what's wrong with that method? where is my mistake?
• Feb 1st 2011, 12:53 PM
mr fantastic
Quote:

Originally Posted by Mppl
There are a set of 500 packages, from this set we know 50 have a defect. In order to inspect the set of packages we select 10 from the 500 packages, testing one by one and putting back the previously tested package before testing another one in order to have always 500 packages to choose, it means, we can test the same package twice or three times, four times,..., ten times.
We reject the set of packages if it happens to test more than 2 packages with defect in our test of 10 packages.

What is the probability of rejecting the set of packages?

I got the right answer using a hypergeometric distribuction, however if I try to use a binomial distribution I don't get the same answer why is that so? with the binomial distribution my procedure was that:

we don't reject the set if we find only zero one or two packages with defect, since we have 50 defective packages in 500 0.1 is the probability of picking up a defective one and 0.9 the probability of picking up a non-defective one then the probability of picking up 0 defective packages is 0.9^10
the prob. of picking up one with defective and 9 nondefective is 0.1*0.9^9 but the one with defect can be the frist the second or the third right? so I have 10 ways of doing that so the probability of having one defective is 10*0.1*0.9^9
and the probability of having two defective is Cn(10,2)*0.1^2*0.9^8
by Cn(10,2) I mean combinations.
suming all the three I have 1-P where P is the prob of rejecting.
the thing is the answer I get back is now right! what's wrong with that method? where is my mistake?

Binomial distribution is only used for cases where the sampling is with replacement (why?)
• Feb 2nd 2011, 03:27 AM
Mppl
well as far as I can see I think thats the case, we have replacement here...
• Feb 2nd 2011, 06:28 PM
awkward
Mppl,

Your procedure sounds correct to me, although 1 would state it more concisely as:

The probability of rejection is

$1 - \sum_{i=0}^2 \binom{10}{i} (0.1)^i \; (0.9)^{10-i}$