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September 16th, 2015, 08:11 AM   #1
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Question How to resolve binomial experiment without binomial theorem.

Hi,

I have the following exercise:
Suppose 20% of the items produced by a factory are defective. Suppose 4 items are chosen at random. Find the probability that: 2 are defective.

This is a binomial experiment, but I don't want to apply directly the Binomial theorem. I would want to resolve with basic probability, precentages, and combinations. I have done some reasonings. But I an unable to finish the exercise correctly.


Let's see the problem in another way:
A factory produces 100 items. 20 items are defective and 80 are not-defective.

4 Items are chosen from the 100 items available,
therefore the ways we can do this selection are: $\displaystyle \binom{100}{4}$

Now, we want to calculate the probability that among these 4 selected items, there can be 2 items defective AND 2 items not-defective. I can select the 2 defective items from 20 defective items in $\displaystyle \binom{20}{2}$ ways, AND I can select the 2 not-defective items from 80 not-defective items in $\displaystyle \binom{80}{2}$. Hence, the probability became:

$\displaystyle P = \frac{\binom{20}{2}\binom{80}{2}}{\binom{100}{4}} = 0.15311$


Now, my book uses binomial theorem with a given result of 0.1536. I have used another method, obtaining a different one 0.15311. Why I have obtained a different result? What's wrong with my reasoning?

Please, can you help me?
Many thanks!
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September 16th, 2015, 12:44 PM   #2
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You started with a finite number (100). The theory is based on an unlimited number.
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September 17th, 2015, 07:36 AM   #3
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Quote:
Originally Posted by mathman View Post
You started with a finite number (100). The theory is based on an unlimited number.
Ok. But, does the exercise can be resolved without using Binomial theorem, or not?
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September 17th, 2015, 01:26 PM   #4
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Quote:
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Ok. But, does the exercise can be resolved without using Binomial theorem, or not?
You can start from first principles, but all you will end up doing is deriving the binomial coefficients for the particular case.
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September 18th, 2015, 09:27 AM   #5
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Quote:
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You can start from first principles, but all you will end up doing is deriving the binomial coefficients for the particular case.
please, can you explain me better?
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September 18th, 2015, 01:38 PM   #6
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A given selection of 4 items with 2 bad and 2 good has a probability of $\displaystyle .8^2.2^2=.0256$

Now count the number of possible combinations with 2 bad and 2 good.
4 items has 4! permutation. However the 2 bad and 2 good are each interchangeable, so that the number of combinations is $\displaystyle \frac{4!}{2!2!}=6$, which is the binomial coefficient for this combination. This leads to .1536.
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