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Math Help - Probability- Explained?

  1. #1
    Len
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    Probability- Explained?

    I was shown how to do this question but I am not sure why it was done like this.

    Four numbers are drawn at random from a box of ten numbers 0,1,...,9.

    Find the probability that the largest number drawn is a six with replacement.

    S={0,1,2,3,4,5,6,7,8,9} #S=10

    A:Set of ordered 4-tuples out of S_1={0,1,2,3,4,5,6} #S_1=7

    B:Set of ordered 4 tuples out of S_2={0,1,2,3,4,5} #S_2=6

    Then,
    P(largest of four numbers drawn with replacement is 6)=
    [#(A)-#(B)]/10^4, =[(7^4)-(6^4)]/10^4

    What I don't understand is why is it [#(A)-#(B)].
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  2. #2
    MHF Contributor matheagle's Avatar
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    I would treat this as a multinomial with 3 groups.
    I would add the probability of exactly 1, exactly 2, exactly 3 or exactly 4 times we select a 6 as the highest number.

     {4 \choose 3,1,0}(.6)^3(.1)^1(.3)^0+{4 \choose 2,2,0}(.6)^2(.1)^2(.3)^0

    +{4 \choose 1,3,0}(.6)^1(.1)^3(.3)^0+{4\choose 0,4,0}(.6)^0(.1)^4(.3)^0

    =.0864+.0216+.0024+.0001=.1105

    The [(7^4)-(6^4)]/10^4 is the different of the two sets.
    If you list all possible sets of four numbers and you first select any number from 0 to 6, but then you remove
    all of those that only consist of 0 through 5 you end up with at least one 6 somewhere.
    Last edited by matheagle; April 8th 2010 at 10:27 PM.
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  3. #3
    Len
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    Quote Originally Posted by matheagle View Post
    I would treat this as a multinomial with 3 groups.
    I would add the probability of exactly 1, exactly 2, exactly 3 or exactly 4 times we select a 6 as the highest number.

     {4 \choose 3,1,0}(.6)^3(.1)^1(.3)^0+{4 \choose 2,2,0}(.6)^2(.1)^2(.3)^0

    +{4 \choose 1,3,0}(.6)^1(.1)^3(.3)^0+{4\choose 0,4,0}(.6)^0(.1)^4(.3)^0

    =.0864+.0216+.0024+.0001=.1105

    The [(7^4)-(6^4)]/10^4 in the different is the two sets.
    If you list all possible sets of four numbers and you first select any number from 0 to 6, but then you remove
    all of those that only consist of 0 through 5 you end up with at least one 6 somewhere.
    Thanks so much, I really appreciate this =)
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