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Math Help - A VERY VERY HARD probability problem.!!!

  1. #1
    Junior Member Kaloda's Avatar
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    Red face A VERY VERY HARD probability problem.!!!

    Inside a box, there are 16 marbles in which 7 are red and the rest are blue. 16 people that consists of 11 males and 5 females lined up to randomly pick one marble at a time. The females occupied the 3rd, 4th, 8th,9th, and 14th spot and the males occupied the rest.

    Assuming that there is no replacement, what is the probability that all female would have red marbles?

    Please HELP!!! Thank you <3.
    Last edited by Kaloda; November 9th 2013 at 11:42 AM.
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  2. #2
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    Re: A VERY VERY HARD probability problem.!!!

    Quote Originally Posted by Kaloda View Post
    Inside a box, there are 16 marbles in which 7 are red and the rest are blue. 16 people that consists of 11 males and 5 females lined up to randomly pick one marble at a time. The females occupied the 3rd, 4th, 8th,9th, and 14th spot and the males occupied the rest. Assuming that there is no replacement, what is the probability that all female would have red marbles?
    If you had a string BBBBBBBBBRRRRRRR how many ways are there to arrange that string?

    Of those rearrangements, how many have an R in the 3rd, 4th, 8th,9th, and 14th spots?
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  3. #3
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    Re: A VERY VERY HARD probability problem.!!!

    The only thing 'hard' about this problem is recognizing that most of the information given is irrelevant. The "a-priori" probability that a given person has a red marble is independent of their place in line. It is sufficient to imagine that the females are first in line. The probability that the first female in line gets a red marble is 7/16, the probability that the second female gets a red marble is (7-1)/(16- 1)= 6/15, the probability the third female gets a red marble is (6- 1)/(15-1)= 5/14, fourth 4/13, fifth 3/12. The probability all 5 females get red marbles is
    \frac{7}{16}\frac{6}{15}\frac{5}{14}\frac{4}{13}\f  rac{4}{13}\frac{3}{12}.
    Last edited by HallsofIvy; November 9th 2013 at 12:16 PM.
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  4. #4
    Junior Member Kaloda's Avatar
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    Re: A VERY VERY HARD probability problem.!!!

    Quote Originally Posted by HallsofIvy View Post
    The only thing 'hard' about this problem is recognizing that most of the information given is irrelevant. The "a-priori" probability that a given person has a red marble is independent of their place in line. It is sufficient to imagine that the females are first in line. The probability that the first female in line gets a red marble is 7/16, the probability that the second female gets a red marble is (7-1)/(16- 1)= 6/15, the probability the third female gets a red marble is (6- 1)/(15-1)= 5/14, fourth 4/13, fifth 3/12. The probability all 5 females get red marbles is
    \frac{7}{16}\frac{6}{15}\frac{5}{14}\frac{4}{13}\f  rac{4}{13}\frac{3}{12}.
    Thanks for the reply but I think your solution is flawed because you completely ignored the fact that there are also male candidates. I mean, what if the the first two males which are ahead of the first female picked the red ones? Won't the probability of "the first female in line" that she will pick a red marble be reduced to 5/14 instead of 7/14 (not 7/16 as you already stated).
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    Re: A VERY VERY HARD probability problem.!!!

    Quote Originally Posted by Kaloda View Post
    Thanks for the reply but I think your solution is flawed because you completely ignored the fact that there are also male candidates. I mean, what if the the first two males which are ahead of the first female picked the red ones? Won't the probability of "the first female in line" that she will pick a red marble be reduced to 5/14 instead of 7/14 (not 7/16 as you already stated).
    No, actually both solutions are correct and equivalent.

    Here are the calculations both.
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  6. #6
    Junior Member Kaloda's Avatar
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    Re: A VERY VERY HARD probability problem.!!!

    Quote Originally Posted by Plato View Post
    No, actually both solutions are correct and equivalent.

    Here are the calculations both.
    Okay. Got it.
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