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Question about a probability problem

  1. Feb 11, 2013 #1

    reenmachine

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    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. jcsd
  3. Feb 11, 2013 #2

    mathman

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    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..
     
  4. Feb 11, 2013 #3

    reenmachine

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    of course!!! I feel a little bit stupid right now , thanks a lot!!!
     
  5. Feb 11, 2013 #4

    CompuChip

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    That doesn't really mean anything, especially in statistics. If you are on the right track, you are usually off by a multiplicative factor instead of a small absolute percentage.

    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 co-incidence (which is quite likely if you mess about with various combinations of 3, 5, 6 and 7 in a fraction).
     
  6. Feb 11, 2013 #5

    reenmachine

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    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 Best-of-7 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...)?
     
    Last edited: Feb 11, 2013
  7. Feb 12, 2013 #6

    CompuChip

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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?
     
  8. Feb 12, 2013 #7

    mathman

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    Be careful - you are confusing two different problems. Picking a ball and not returning is different from best of seven for sports. The latter is analogous to picking a ball and then returning it before picking the next ball.
     
  9. Feb 13, 2013 #8

    reenmachine

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    Suppose you have 2 fantasy teams and 23 judges who votes on the series , then the result comes out 16 and 7 and we want to decide who wins game 1 , game 2 and so on until one team wins 4 out of 7 games , wouldn't my technic to pick a ball and not return it to the bag be the way to go? If so , that's why I asked how to calculate the number of possible ways wbbwbwb wwwwbbb etc...

    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?
     
    Last edited: Feb 13, 2013
  10. Feb 13, 2013 #9

    reenmachine

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    I'm not going to school , but I'll try to find the explanation online and self-teach it to myself if possible.

    thanks
     
  11. Feb 13, 2013 #10

    mathman

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    I think you should describe this scenerio in more detail. What happens in game 1, what happens in game 2, etc.? What is the 16 - 7 division in reference to?
     
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