1. The problem statement, all variables and given/known data In a company, out of a sample of 20 bulbs, the mean number of defective bulbs are 2. Out of 1000 such samples, how many samples would have atleast 3 defective bulbs? 2. Relevant equations Mean = n * p where n is number of bulbs and p is probability of a bulb being defective. P(x succeed at r times) when the experiment is done n times = nCr prq(n-r) where p is probability success when experiment is done once and q = 1 - p 3. The attempt at a solution Mean number of defectives = 2 = n * p. n = 20. so p =2/20 = 0.1. Probability of bulb being defective and q = 1 - 0.1 = 0.9. Probability of bulb not being defective. Probability of at least 3 defective in a sample of 20 = 1 - [ P(zero defective) + P(1 defective) + (P(2 defective) ] Now n = 20. r = 0 here r =1 here r = 2 here p = 0.1 q = 0.9 P(atleast 3 in sample of 20) = 1 - [ 20C0 (0.1)0(0.9)20 + 20C1(0.1)1 (0.9)19 + 20C2(0.1)2(0.9)18 ) This i get as 0.323. So the probability of at least three defective in a sample of 20 bulbs is 0.323. Now how to proceed? I have to find 'How many such samples would be expected to contain atleast 3 defective bulbs'? Answer is 323. I guess they've multplied 0.323 by number of samples but why? PS... sorry for too many homework posts. Have an exam coming on 11th Feb.