Hypergeometric Distribution Problem

#1
Looking for possible solutions to the problem below:
"A particular supplier guarantees a quality level of 99%. You recently purchased 100 components. You have built 50 subassemblies which require 1 component each. In the 50 subassemblies, you found 3 defective components. What are the chances that this could happen randomly."
 
#2
Just to provide some background and to show I am not just looking for a pass on a "homework":)...I ran in to this problem in a Six sigma BB class and I was not satisfied with the solution presented. This would have been a direct substitution to the formula provided:

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However K is not provided. Instead it was given that the quality level guaranteed was @99%. The solution presented basically assumed K = 1:
reject rate = 1 - 99% = 1%, 1% of a 100 is 1
This results into having a C( 1 , 3) in the calculation so the result is 0.

We had just covered binomial distributions prior to this topic, so naturally, I questioned this assumption. My idea is you have to consider all scenarios that could have occurred when the 100 components were purchased, which includes having zero to 100 defects in the population. The probability for each scenario will need to be calculated and multiplied to the probability for the hypergeometric case associated with it. These products would then need to be totaled to get the overall probability:

ie. for the case 3 of 100 defective: Binomial distribution probability is 6.1%, Hypergeometric probability given k=3 is 12.1%, giving us 0.74%

note: case for K= 0,1,2 and k>53 are excluded as it is impossible to get exactly 3 defectives out of the 50 samples in these cases.
 

katxt

Active Member
#5
The probability for each scenario will need to be calculated and multiplied to the probability for the hypergeometric case associated with it. These products would then need to be totaled to get the overall probability:
I'm pretty sure that if you go from K = 3 to 100 with N = 100, (or 53 as Ynon shows) calculate the probability of getting 3 out 50 for each K using the hypergeometric, then find the probability of getting each K using binomial with N = 100 and p = 0.01, then multiplying and adding as Ynon suggests, you will end up with the binomial probability P(3/50 where p of each is 0.01) as I suggested a couple of posts back.
 
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#7
First of all thanks for the responses guys:)
I got the exact same results as katxt. I first approached it as a hypergeometric as it was specified as a hypergeometric sample problem (had to apply the concept:)). But upon reflection on the problem I realized it didn't matter if the 50 units came from the 100 samples since we had no info on the actual defect count in that sample. So I agree, it is just a binomial distribution with sample 50 and 1% p. I couldn't explain how the results were exactly the same though I was expecting them to be very close. Maybe you can shed light on that as well?

Here's another part of the story, I reached out to the guy who created the problem (basically a Six Sigma guru from Motorola). He insists the 1% was not to be treated as a population parameter but a sample statistic for the 100 units. I contested this since it wasn't realistic for a supplier to declare or guarantee defect rates for a certain batch (in this case the 100 units). Furthermore, if we assume 1 defect from the 100, we already know the probability is zero to get 3 defects from the subset of 50 (no need to do a hypergeometric). What do you guys think?