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Combinatorics...Please help

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data
    There are 20 eggs in the box and three of them are rotten,

    If I pick eggs without replacement calculate P(my first rotten egg comes at my k-th pick)
    2. Relevant equations


    3. The attempt at a solution
    I come up with two solutions:
    (1)As there is no replacement,I can use the hypergeometric distribution
    So,P(my first rotten egg comes at my k-th pick) = C(17,k-1)C(3,1) / C(20,k)

    (2)I can use the multiplication theorem for conditional probability,i.e.
    P(my first rotten egg comes at my k-th pick) = (17/20)(16/19).....(3/21-k)

    Let assume I got a rotten egg in the 4-th trial,
    Using method (1),P = 8/19
    Using method (2),P = 2/19

    Similarly,for the 5-th trial
    (1) yields P = 35/76
    (2) yields P = 7/76

    In each case,(1)/(2) = k,where does this factor k come from?Both of them seems make sense so which one is wrong?

    Thanks :)
     
  2. jcsd
  3. Nov 3, 2015 #2

    Samy_A

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    Sorry, I was mistaken.
     
  4. Nov 3, 2015 #3
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    Last edited: Nov 3, 2015
  5. Nov 3, 2015 #4
    Sorry for keep editing my post,my thought isn't clear.
    Do you mean method(2)?
    The probability of choosing good eggs out of 20 is 17/20
    out of 19 is 16/19
    .
    .
    .
    Then suppose you got a rotten at the 4-th trial,
    the probability would be (17/20)(16/19)(15/18)(3/17),because there are 3 rotten eggs out of 17 eggs
    If we generalize the above ,we have
    1 rotten egg = 3/20
    2 rotten eggs = (17/20)(3/19)
    3 rotten eggs = (17/20)(16/19)(3/18)
    k rotten eggs = (17/20)........(3/21-k)
     
  6. Nov 3, 2015 #5

    Samy_A

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    Yes, I was mistaken, sorry for that.

    Your first method is wrong, because in the nominator you multiply the number of ways to select (k-1) non rotten eggs with 3 to get the number of positive cases. But that is clearly wrong, as the number of ways to get the first rotten egg at the k-th pick must be less that the number of ways to select (k-1) non rotten eggs.

    Your second method seems right, although there is something wrong with the indices in the product (edit: you removed the product, so forget this remark :oldsmile:).
     
  7. Nov 3, 2015 #6
    Do you mean the hypergeometric distribution would only make sense if we are choosing several objects in 1 trial?So in this case,in which the eggs are picked one by one,the hypergeometric is no longer applicable?
     
  8. Nov 3, 2015 #7

    Samy_A

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    The hypergeometric distribution gives the probability of k successes in n draws (at what place in the picking order the successes occur is irrelevant).
    That is not what is asked here: you are asked what the probability is that the k-th draw is the first success.
     
  9. Nov 3, 2015 #8

    Ray Vickson

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    Your calculations are correct, but are severely mis-labelled. You have three rotten eggs---period. You are asked to find the probability of encountering your first rotten egg on draw k, so if ##X =## draw number of first rotten egg, then you should, indeed, have
    [tex] P(X = 1) = \frac{3}{20}, \: P(X = 2) = \frac{17}{20} \frac{3}{19}, \; \text{etc.} [/tex]

    You could have labelled these instead as P(0 good eggs) = 3/20, P(1 good egg) = (17/20)(3/19), etc., because you would be counting the "good" eggs before your first rotten one.

    BTW: this is not a "combinatorics" problem; it is a "probability" problem.
     
  10. Nov 3, 2015 #9
    Thx for reminding, I always have a problem on defining the RV(i.e.X=1,X=2) so I usually just do it in that way, though it is one of the culprits lead to confusions. I will write them more precisely.
     
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