a buyer will accept a lot of 100 articles if a sample of 5 picked at random and inspected contains no defects. What is the probability that the buyer will accept the lot if it contains 10 defective articles?
I have tried binomial formula, decision trees and nothing works to arrive at the correct answer of .58. The closest result I got was (.095)/(.95)(.1)+(.9)(.05))=.68 , but it's still not the correct answer.
Assume that you don't put back the ones that have already been picked. Then, the probability of picking 5 good ones in a row is: $$ \left(\frac{90}{100}\right)\left(\frac{89}{99}\right)\left(\frac{88}{98}\right)\left(\frac{87}{97}\right)\left(\frac{86}{96}\right)=0.583752. $$
Alternatively, there are $\binom{90}{5}$ ways to pick 5 out of 90 good ones. Divide this by $\binom{100}{5}$, the number of ways to pick 5 out of 100, will also yield $0.583752$.