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Math Help - chance of picking colors in pairs.

  1. #1
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    Unhappy chance of picking colors in pairs.

    Hi All!

    I have 10 black balls and 12 white balls (mixed).
    The balls are randomly sorted to pairs.

    Q: what is the chance of picking 5 pairs with white and black balls in each and 1 pair of 2 remaining white balls?


    I don't have a clue how to begin solving it.
    Hope you can help.

    Thank you so much in advance!
    Danny
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  2. #2
    Pim
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    Well, I'm not entirely sure, but I do have an idea how it could be done.

    What you do is fix the first ball of each pair, then "move" the other ones
    (Drawing with O as black and X as white)

    |O|O|O|O|O|X|X|X|X|X|X|
    | | | | | | | | | | | |

    Now, the leftover balls can be sorted in

     <br />
\frac{11!}{5!6!}<br />
ways.
    only 1 of these is correct. (Namely with first all black balls then all white ones.)

    Therefore the probability is
     <br />
\frac{1}{\frac{11!}{5!6!}}<br />

    which comes down to

    \frac{1}{462}

    which seems about right to me. (Again, I'm not 100% positive this is the correct way, but the outcome seems to be right.)
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  3. #3
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    Probability

    Hello Danny
    Quote Originally Posted by Dindia View Post
    Hi All!

    I have 10 black balls and 12 white balls (mixed).
    The balls are randomly sorted to pairs.

    Q: what is the chance of picking 5 pairs with white and black balls in each and 1 pair of 2 remaining white balls?


    I don't have a clue how to begin solving it.
    Hope you can help.

    Thank you so much in advance!
    Danny
    I have to choose a total of 12 balls from the 22 original balls, one at a time, without replacing each choice.

    The probability that my first choice is a white ball is \frac{12}{22}. If this ball is white, the probability that my second choice is black is \frac{10}{21}.

    So the probability that I choose white followed by black is \frac{12}{22}\times\frac{10}{21}.

    But we want the first pair to be one of each colour. So we must also allow the choice to be black followed by white. This is \frac{10}{22}\times \frac{12}{21}, by a similar argument. You'll see that this is the same fraction as before. So the probability that the first pair I choose contains a white ball and a black in either order is \frac{2.12.10}{22.21}.

    If the first pair does contain one of each colour, then there are 11 white and 9 black remaining. So, in the same way as before, the probability that the second pair I choose contains a ball of each colour is \frac{2.11.9}{20.19}.

    Continue in the same way for the 3rd, 4th and 5th pairs.

    The sixth pair is different. By now, if all the previous events have occurred, there are 7 white and 5 black balls left. The probability that the next two chosen are both white is \frac{7.6}{12.11}.

    Multiply all these six probabilities together to obtain the overall probability of all these events occurring. Expressed as factorials, this comes to \frac{2^5.12!.(10!)^2}{(5!)^2.22!}.

    Or, if you're careful you can cancel the numbers to get an exact answer, which is \frac{576}{46,189} \approx 0.0125.

    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello Danny

    I have to choose a total of 12 balls from the 22 original balls, one at a time, without replacing each choice.

    The probability that my first choice is a white ball is \frac{12}{22}. If this ball is white, the probability that my second choice is black is \frac{10}{21}.

    So the probability that I choose white followed by black is \frac{12}{22}\times\frac{10}{21}.

    But we want the first pair to be one of each colour. So we must also allow the choice to be black followed by white. This is \frac{10}{22}\times \frac{12}{21}, by a similar argument. You'll see that this is the same fraction as before. So the probability that the first pair I choose contains a white ball and a black in either order is \frac{2.12.10}{22.21}.

    If the first pair does contain one of each colour, then there are 11 white and 9 black remaining. So, in the same way as before, the probability that the second pair I choose contains a ball of each colour is \frac{2.11.9}{20.19}.

    Continue in the same way for the 3rd, 4th and 5th pairs.

    The sixth pair is different. By now, if all the previous events have occurred, there are 7 white and 5 black balls left. The probability that the next two chosen are both white is \frac{7.6}{12.11}.

    Multiply all these six probabilities together to obtain the overall probability of all these events occurring. Expressed as factorials, this comes to \frac{2^5.12!.(10!)^2}{(5!)^2.22!}.

    Or, if you're careful you can cancel the numbers to get an exact answer, which is \frac{576}{46,189} \approx 0.0125.

    Grandad
    Hi Grandad,

    I agree with your analysis up to a point, but it assumes the white-white pair is the last pair taken. I think the problem statement, on the other hand, allows it to be any one of the 6 pairs. Each of these 6 outcomes has the same probability, so I think we need to multiply your answer by 6, for a final probability of 0.074823.
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  5. #5
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    Probability

    Hi Danny
    Quote Originally Posted by awkward View Post
    Hi Grandad,

    I agree with your analysis up to a point, but it assumes the white-white pair is the last pair taken. I think the problem statement, on the other hand, allows it to be any one of the 6 pairs. Each of these 6 outcomes has the same probability, so I think we need to multiply your answer by 6, for a final probability of 0.074823.
    I think you could well be right. I certainly assumed from the wording of the question - especially from the use of the word 'remaining' - that the white-white pair was the last pair to be revealed. But if it can come anywhere in the order, then you're right to multiply by 6.

    Grandad
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  6. #6
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    Hi Grandad,

    The problem statement is, I concede, unclear or at least ... awkward.
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