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Math Help - marbles

  1. #1
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    marbles

    A bag contains 4 red marbles and 2 blue marbles. You draw a marble at random, without replacement, until the first blue marble is drawn. Let X = the number of draws.

    a) If you repeated this experiment a very large number of times, on average how many draws would you make before a blue marble was drawn?
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  2. #2
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    Quote Originally Posted by swoopesjr01
    A bag contains 4 red marbles and 2 blue marbles. You draw a marble at random, without replacement, until the first blue marble is drawn. Let X = the number of draws.

    a) If you repeated this experiment a very large number of times, on average how many draws would you make before a blue marble was drawn?
    Not sure about this one. I guess that one which is most likely to happen.

    Probability
    On 1st]=2/6=1/3=.3333
    On 2nd]=(4/6)(2/5)=4/15=.2666
    On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
    On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
    On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

    It seems like on the first try to me.
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  3. #3
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    the answer is = 2.334..im not really sure i understand what you did in your reply
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    Not sure about this one. I guess that one which is most likely to happen.

    Probability
    On 1st]=2/6=1/3=.3333
    On 2nd]=(4/6)(2/5)=4/15=.2666
    On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
    On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
    On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

    It seems like on the first try to me.
    The average number of draws is the mean \sum_{x=1}^{5} xf(x) where f(x) are PH's probabilities. The mean is 2.33 as swoopes says.

    Why don't quotes within quotes work? Using the quote button on PH's post loses his quote of swoopes.
    Last edited by JakeD; May 9th 2006 at 10:32 PM.
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  5. #5
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    Quote Originally Posted by JakeD
    Quote Originally Posted by PH

    Originally Posted by ThePerfectHacker
    Not sure about this one. I guess that one which is most likely to happen.

    Probability
    On 1st]=2/6=1/3=.3333
    On 2nd]=(4/6)(2/5)=4/15=.2666
    On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
    On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
    On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

    It seems like on the first try to me.
    The average number of draws is the mean \sum_{x=1}^{5} xf(x) where f(x) are PH's probabilities. The mean is 2.33 as swoopes says.

    Why don't quotes within quotes work? Using the quote button on PH's post loses his quote of swoopes.
    They do, you just have to include them yourself

    RonL
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  6. #6
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    Quote Originally Posted by CaptainBlack
    Originally Posted by JakeD
    Quote:
    Originally Posted by PH

    Originally Posted by ThePerfectHacker
    Not sure about this one. I guess that one which is most likely to happen.

    Probability
    On 1st]=2/6=1/3=.3333
    On 2nd]=(4/6)(2/5)=4/15=.2666
    On 3rd]=(4/6)(3/5)(2/4)=2/10=.2000
    On 4th]=(4/6)(3/5)(2/4)(2/3)=2/15=.1333
    On 5th]=(4/6)(3/5)(2/4)(1/3)(2/2)=1/15=.0666

    It seems like on the first try to me.





    The average number of draws is the mean where are PH's probabilities. The mean is 2.33 as swoopes says.

    Why don't quotes within quotes work? Using the quote button on PH's post loses his quote of swoopes.

    They do, you just have to include them yourself

    RonL
    But you may have some trouble includong the TeX (unless you have moderator privaleges )

    RonL
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  7. #7
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    what would be the probability of obtaining the same blue marble on the 6th draw if it was done with replacement?
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