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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.

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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