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Question about a probability problemby reenmachine
Tags: probability 
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#1
Feb1113, 02:55 PM

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P: 521

Hi all , here's hoping someone could help me with this problem. :)
The problem is basically as follow: You have 7 balls in your pocket , 3 are whites and 4 are blacks.If you take out 3 balls out of your pocket , what are the odds that you'll have 2 white balls and 1 black ball in your hand ? Now I have no mathematic education at all so here's how I proceeded: 3/7 * (1/2 + 1/3) * (1/5 + 2/5 + 3/5) 3/7 * 5/6 * 5/5 3/7 * 5/6 = 5/14 5/14 = 35,7% = 36% That was my final answer and I got it wrong , the answer happened to be 34%. How could my method of calculating the odds produce an answer that was so close yet wrong? Help would be greatly appreciated :D thanks in advance 


#2
Feb1113, 03:28 PM

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P: 6,077

There are three ways to get the desired result. wwb, wbw, bww.
P(wwb) = (3/7)(2/6)(4/5) P(wbw) = (3/7)(4/6)(2/5) P(bww) = (4/7)(3/6)(2/5) Each term = 4/35. Sum = 12/35 = .342857.. 


#3
Feb1113, 03:35 PM

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P: 521

of course!!! I feel a little bit stupid right now , thanks a lot!!!



#4
Feb1113, 03:51 PM

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Question about a probability problem
For example, if you'd done the P(wwb) correctly but forgot to take into account the other two possibilities you would be off by exactly a factor of 3. That you get 36% instead of 34% (after rounding both numbers to integer percentages) is probably just a coincidence (which is quite likely if you mess about with various combinations of 3, 5, 6 and 7 in a fraction). 


#5
Feb1113, 04:13 PM

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P: 521

Another question related to this problem , finding all the ways to get there (wwb,wbw,bww) is pretty easy in this case , but suppose the problem would require me to find all the possible ways for a larger sample like for example:
I have 7 white balls and 16 black balls in a bag (each ball indicate a potential victory for each team) and I want to calculate the odds of winning in a Bestof7 sport series for the white team (of course each time a ball is picked she doesn't return in the bag) is there a quicker way to do it than trying to write all the ways down (like wwwwbbb , wwwbwbb etc...)? 


#6
Feb1213, 01:58 PM

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Yes, there is an easy way of doing that, it's called the binomial coefficient:
[tex]\binom{16}{7} = \frac{16!}{7! 9!}[/tex] Before we start writing a complete Wikipedia article here, don't you have any textbook or course material that explains this? 


#7
Feb1213, 03:30 PM

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#8
Feb1313, 10:54 AM

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P: 521

Or is what you are saying is that I should always keep all the balls in the bag before each game , making it a 16/7 odds for every game and ultimately the series outcome? 


#9
Feb1313, 10:55 AM

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P: 521

thanks 


#10
Feb1313, 03:29 PM

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P: 6,077




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