Probability (binomial Distro?)

[probabilityst]

Hi Everyone, I've been having some trouble with this problem:

A lot of 250 items that contains 15 defective items is subject to an acceptance sampling plan that calls for a smple of size 6 to be drawn(without replacement). the lot is accepted if the sample contains, at most, one defective item. find the probability that the lot is accepted.


The way I thought to do it was:
total number of samples that would be able to pass/total number of sample possible = (250-14)C6 / 250C6.

A friend of mine said that it should be done like this: (6c0) * 0.06^0 * (0.94)^6 + 6c1 * (0.06^1) * (0.94)^5, and something about it being a binomial distribution, but I don't know why/what he's saying.

Any help would be greatly appreciated.

 

[lanedance]

note the binomial assumes a constant probabilty of success (failure). As p will effectively change as failed items are removed from the universe without replacement it is not exactly right but will be very close.

see binomial distribution http://en.wikipedia.org/wiki/Binomial_distribution
P(k success in n trials with average probability p) = f(n,k,p)

here the binomial approximation is
p = 15/250, failure rate
n = 6, samples
k = 0 or 1 observed failures

then the probabilty that the sample is accepted is
P(lot failures <=1) = f(0,6,p) + f(1,6,p)

note
f(0,6,p) is just the proability of getting no failures from 6 selected items with constant p
f(1,6,p) is just the proability of getting 1 failure from 6 selected items with constant p

I would calculate the probabilties by hand assuming no replacement and compare against the binomial case

 

[LCKurtz]

Hi Everyone, I've been having some trouble with this problem:

A lot of 250 items that contains 15 defective items is subject to an acceptance sampling plan that calls for a smple of size 6 to be drawn(without replacement). the lot is accepted if the sample contains, at most, one defective item. find the probability that the lot is accepted.


The way I thought to do it was:
total number of samples that would be able to pass/total number of sample possible = (250-14)C6 / 250C6.

A friend of mine said that it should be done like this: (6c0) * 0.06^0 * (0.94)^6 + 6c1 * (0.06^1) * (0.94)^5, and something about it being a binomial distribution, but I don't know why/what he's saying.

Any help would be greatly appreciated.

The appropriate distribution is the hypergeometric. See, for example,

http://en.wikipedia.org/wiki/Hypergeometric_distribution

 

[lanedance]

nice point, couldn't remember the distribution