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Probability Question

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Positive integers i, j, k, m, and n are randomly chosen (repetition is allowed) so that
    2 <= i,j,k,m,n <= 2009 . What is the probability that ijk + mn is even?


    2. Relevant equations

    3. The attempt at a solution

    Each of the positive integers is one out of 2008 numbers. 1004 of these numbers are even and 1004 of these numbers are odd. So, 50% chance of being an even number and 50% chance of being an odd number for each of the integers i,j,k,m,n.

    I looked at the ijk term first and wrote out all possible combinations.

    odd x odd x odd = odd
    odd x odd x even = even
    odd x even x odd = even
    odd x even x even = even
    even x odd x odd = even
    even x odd x even = even
    even x even x odd = even
    even x even x even = even

    1/8 possibilities are odd (12.5%) and 7/8 are even (87.5%)

    I looked at the mn term next and did the same.

    odd x odd = odd
    odd x even = even
    even x even = even
    even x odd = even

    1/4 possibilities are odd (25%) and 3/4 are even (75%)

    Next, I looked at the term ijk and the term mn added together.

    odd + odd = even
    odd + even = odd
    even + odd = odd
    even + even = even

    2/4 possibilities are odd (50%) and 2/4 are even (50%)


    I'm really not sure where to go from here. Am I on the right track at least? I'm just not sure how to combine the above facts into a statement about the probability of [ijk + mn] being an even number. Any help is much appreciated :smile:
     
  2. jcsd
  3. Nov 11, 2009 #2
    Um, I might be wrong here, long since my last prob class but I believe you're almost there:
    For ijk +mn to be even IJK AND MN must be either odd or even, so why not calculate the probability of BOTH being odd (intersection between the chances of ikj being odd and mn being odd), and the probability of both being even (prob of ijk being even intersection mn being even), and then compute the union of both probability sets.

    I might be wrong, but i hope i can be of help
     
  4. Nov 11, 2009 #3
    Chances of ijk being odd = 12.5%
    Chances of mn being odd = 25%
    To calculate the intersection for the above set would it just be (1/8)x(1/4) = (1/32) or 3.125%?


    Chances of ijk being even = 87.5%
    Chances of mn being even = 75%
    And likewise, for the above set would it be (7/8)x(3/4) = 21/32 or 65.625%?


    I'm not sure what you mean by the union of the two sets. How would the union between 3.125% and 65.625% be calculated? I'm not in a probability and stats class so sorry if this is an elementary question :)
     
  5. Nov 11, 2009 #4
    The union of two probability sets is the probability of A happening OR B happening:
    The chances of throwing a dice and getting 6 is 1/6, the chance of getting a 2 is 1/6, the chance of getting a 6 or a 2 is 1/6 + 1/6 = 1/3.

    The intersection of two probability sets, which is the probability of A and B happening is computed like this:
    The chances of getting a 6 is 1/6, the chances of getting a 2 is 1/6, the chances of throwing a 6 and then throwing a 2 are 1/6 x 1/6 = 1/36.
     
  6. Nov 11, 2009 #5
    I understand now, thank you for your help.
     
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