Hypergeometric Probability DistributionDate: 08/08/99 at 14:04:22 From: Cindi Ernsdorff Subject: Probability and Statistics Problem: A box contains 3 defective light bulbs and 7 good light bulbs. Three consecutive bulbs are drawn at random from the box (they are not replaced after being drawn; this means that this is not a Bernoulli trial experiment.) A) Construct a tree to show all of the possible outcomes. B) What is the probability that 3 bad bulbs will be drawn? I have figured out how to draw a tree if there WERE repetitions, 3/10 on the top, and 7/10 on the bottom branch, with each branch consisting of those two 'pieces', and then doing them 3 times each. I cannot figure out how to do NON-repetitions, nor how to figure out how to construct the tree. Please help! Thank you, Cindi Date: 08/08/99 at 17:26:28 From: Doctor Anthony Subject: Re: Probability and Statistics This is a problem using the hypergeometric probability distribution. I shall be using the C(n,r) notation as described below. C(n,r) Explanation ------------------- Use the notation C(n,r) to mean the number of combinations of r things that can be made from n different things. The formula for C(n,r) is: n! 10! C(n,r) = -------- so C(10,4) = ----- = 210 r!(n-r)! 4! 6! Here we have 10 bulbs, of which 3 are defective and 7 good. If 3 are chosen at random, then: C(3,3) 1 Probability all 3 are defective = ------- = --- C(10,3) 120 The tree diagram would look like this 1st Draw Defective | Good |-------------------------| | | 3/10 7/10 | | |-----|------| |------|------| 2nd draw | | | | 2/9 7/9 3/9 6/9 | | | | |-----------| |--------| |----------| |----------| 3rd draw | | | | | | | | 1/8 7/8 2/8 6/8 2/8 6/8 3/8 5/8 DDD DDG DGD DGG GDD GDG GGD GGG 1/120 7/120 7/120 7/40 7/120 7/40 7/40 7/24 = Prob. where D = defective, G = good You multiply the branch probabilities to find the probability of ending up with a particular outcome. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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