A box contains ten assemblies of which two are defective. A sample of three assemblies is selected at random. What is the probability that the two defective parts will be selected? A.0.0667 B.0.0776 C.0.667 D.0.433 E.0.20 Why is my colculation not correct? Where am I wrong? Probability of Defective : 0.2 Probability of Non Defective : 0.8 Selecting 2 defective ones: 1/5*1/5*4/5 =4/125 OA is A

Hi, Both the defensive ones are to be picked.This can be done in 1 way From the remaining 8 one has to be picked in 8C1 ways. Total number of ways of picked 3 out of 10 is 10C3 So, probability = 1*8C1/10C3 = 8/120 = 1/15 Hence, A Coming to your method. 1st defective is selected in 2/10 ways After it is picked 2nd one is picked in 1/9 ways Third one is picked in 8/8 ways So, probability is 1/5*1/9*1 = 1/45 But, here we are considering the order. In this case, the non-defective one is picked 3rd. It can be picked first or second as well When non-defective one is picked first, probability is (8/10)*(2/9)*(1/8) = 1/45 When non-defective one is picked second, probability is (2/10)*(8/9)*(1/8) = 1/45 So, probability is 3(1/45) = 1/15.

Answer is A. The answer should be solved in this way: You have in all 10C3 choices that is 120 choices, so this is your denominator. In the Numerator you have 2 fixed choices to get the 2 defectives so 2C2 * 8C1(which is any assembly out of the remaining 8) Hence, 8/120 = 0.0667