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Probability question - hypergeometric distribution?

  1. Aug 30, 2014 #1
    Hi,

    I have never quite worked this type of probability question out, so would like some help please.

    Imagine this scenario:
    There are 4 people sat around a table, A, B, C and D.
    A is sitting opposite C, B is sitting opposite D.
    There is a bag with 16 balls numbered 1-16.
    The balls are randomly drawn so that each person has 4 balls.

    Imagine you are A.
    And the balls you have are 1,4,5,9.

    Now you want to work out what is the probability that C has a ball that is higher than your highest ball i.e. you want to work out whether C has AT LEAST one of 10,11,12,13,14,15, or 16.

    Now the way I look at it is that the:
    probability of C having ball 10 is 1/3
    probability of C having ball 11 is 1/3
    probability of C having ball 12 is 1/3
    probability of C having ball 13 is 1/3
    probability of C having ball 14 is 1/3
    probability of C having ball 15 is 1/3
    probability of C having ball 16 is 1/3

    But I just don't know what to do next. Any assistance would be appreciated.

    Thanks.
     
  2. jcsd
  3. Aug 30, 2014 #2

    Matterwave

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    Well I would look at it this way. There are 7 balls that beat your 9, and 5 balls that lose. So if C is picking 4 balls at random from the other 12 balls available, what are his chances of not picking a ball that is greater than 9?
     
  4. Aug 31, 2014 #3
    Ah. Ok. So basically you're saying that of the remaining 12 balls, the probability that C has at least one ball higher than my highest ball which is 9 is (1 - probability that C has all balls lower than mine).

    Which comes to

    1 - (5/12 * 4/11 * 3/10 * 2/9) = 1 - 120/11880 = 0.98...

    That seems rather high to me, am I making a silly mistake somewhere?

    Thanks.
     
  5. Sep 1, 2014 #4
    You are making a silly mistake thinking it is high - look at the balls that are left and think about which combinations C could have that don't have any balls above 9,
     
  6. Sep 1, 2014 #5

    Matterwave

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    This is correct. Of the remaining balls, there is only 5 balls that are lower than 9. If C picks 4 random balls out of 12, it's hard for him to always pick out of those 5 which are lower than 9 and get none of the 7 balls that are higher.
     
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