PDA

View Full Version : probability: 20 defective in 500; prob. all 10 defective



anasylum
03-02-2008, 11:30 AM
A container with 500 devices has 20 defective ones. If the devices are selected together from a container, what's the probability that all 10 are defective?

ok, assuming the equation might looks like this 500!/490!10!, can anyone help? I can cancel out the 490!, and reduce some numbers but I still find what i have left to be too high, anyone with any thoughts?

galactus
03-02-2008, 12:21 PM
This is a hypergeometric distribution. Like a binomial only without replacement.

Given a population of N items having k successes and N-k failures, the probability of selecting a sample of size n that has x successes and n-x failures is given by:

\frac{C(k,x)C(N-k,n-x)}{C(N,n)}

Can you finish using the formula?.

anasylum
03-02-2008, 02:57 PM
Hi,
Not sure yet. What would the C represent?

galactus
03-02-2008, 03:00 PM
You're not familiar with combinations?. That just the notation for combinations. How many ways to choose k items from n items when order does not matter is C(n,k).

anasylum
03-02-2008, 03:23 PM
Hi, I am somewhat familiar with combinations and permutations, but this questions is stumping me. Sorry, brain dead here. I'm simply at a loss. Thanks for providing the formula, I just can't work it, but I will keep trying.

All the best,

galactus
03-02-2008, 03:57 PM
Well, you have N=500 and there are 20 defects in that 500. That means 480 are good.

You are drawing 10 from the 500 and finding the probability that all of those are defects. So, you are choosing 0 of the 480 good ones and 10 of the 20 bad ones. Intuitively, I would say the probability is mighty low.

So, you have(from the formula):

\frac{C(480,0)C(20,10)}{C(500,10)}