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:

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.