Question 32804
just think about all the different permutations here: the question is asking for the probability of there being 1 or 2 or 3 or 4 or 5 defectives in his sample: call this P(A). This is very complicated to do. Well, not complicated as susch, just time consuming and boring as Hell after about 3 minutes!


However, we can use the fact that P(A) = 1 - P(not A) to make our work simpler.


If we can find the probability that the inspector picks 5 pieces that are not defective and then take this away from 1, we will have our answer.


So, P(all 5 are not defective) = (27/35)*(26/34)*(25/33)*(24/32)*(23/31)
P(all 5 are not defective) = 0.24868


So, P(at least one defective) = 1 - 0.24868
P(at least one defective) = 0.75132


jon.