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Hypergeometric Probability Distribution


Date: 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/   
    
Associated Topics:
High School Probability

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