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September 16th, 2015, 08:11 AM  #1 
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  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 notdefective. 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 notdefective. I can select the 2 defective items from 20 defective items in $\displaystyle \binom{20}{2}$ ways, AND I can select the 2 notdefective items from 80 notdefective 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! 
September 16th, 2015, 12:44 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,214 Thanks: 492 
You started with a finite number (100). The theory is based on an unlimited number.

September 17th, 2015, 07:36 AM  #3 
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  
September 17th, 2015, 01:26 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,214 Thanks: 492  
September 18th, 2015, 09:27 AM  #5 
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  
September 18th, 2015, 01:38 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,214 Thanks: 492 
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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