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Math Help - Probability of getting 2 pairs

  1. #1
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    Probability of getting 2 pairs

    Hello

    You are dealt 4 cards (each card is numbered from 0 - 9). Each card is independed of the other card. What is the probability of getting 2 pairs?

    P(2 pair) = ­4C2P(1st = 2nd)P(3rd ≠ 2nd = 1st) P(4th = 3rd ≠ 2nd = 1st) = (6)(0.1)(0.9)(0.1)=0.054
    Is this correct?
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  2. #2
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    You've doubled it.
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  3. #3
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    So it's supposed to be:

    P(2 pair) = (0.5)(4 Choose 2)(0.1)(0.9)(0.1)= 0.027?

    If so, why would that be?

    Is it because I would be counting (for example) 1212 twice?
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  4. #4
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    Yes, once with the ones chosen by (4 choose 2)(0.1) and the twos by (0.9)(0.1), and once again the other way.
    Last edited by tom@ballooncalculus; March 10th 2011 at 06:02 AM. Reason: (0.1)
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  5. #5
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    Thank you for your help. Just one more thing, did I make a mistake in the following calculations:

    P(4 different digits) = P(2nd ≠1st)P(3rd ≠1st ≠2nd) P(4th ≠ 2nd ≠ 3rd ≠ 1st) = (0.9)(0.8)(0.7) = 0.504

    P(pair) = (4 Choose 2)P(1st = 2nd)P(3rd ≠ 2nd = 1st)P(4th ≠ 3rd ≠ 2nd = 1st) = (6)(0.1)(0.9)(0.8) = 0.432

    P(Three-of-a-kind) =(4 Choose 3)P(1st = 2nd = 3rd ≠ 4th) = (4)(0.1)2(0.9) = 0.036

    P(Four-of-a-kind) = P(1st = 2nd =3rd =4th ) = 0.001
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  6. #6
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    No, those are correct.
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  7. #7
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    Thank you for all your help.
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