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Math Help - Probability Function

  1. #1
    da`
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    Probability Function

    Struggling here with this question would really appreciate and be very thankful for any help. It seems really general to me and clearly don't have a knack for this, and just need to understand how create a probability function.

    Quote:
    Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.

    Find the probability function of X and the probability function of Y, if sampling is without replacement.

    Find the probability function of X and the probability function of Y, if sampling is with replacement.
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  2. #2
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    Quote Originally Posted by da` View Post
    Struggling here with this question would really appreciate and be very thankful for any help. It seems really general to me and clearly don't have a knack for this, and just need to understand how create a probability function.

    Quote:
    Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.

    Find the probability function of X and the probability function of Y, if sampling is without replacement.

    Find the probability function of X and the probability function of Y, if sampling is with replacement.
    Some things to think about when finding the distribution for X:

    What event gives X = 0? So what's Pr(X = 0) ....
    What events give X = 1? So what's Pr(X = 1) ....
    What events give X = 2? So what's Pr(X = 2) ....
    .
    .
    .
    What events give X = 9? So what's Pr(X = 9) ....


    Things to think about when finding the distribution for Y:

    What event gives Y = 0? So what's Pr(Y = 0) ....
    What events give Y = 1? So what's Pr(Y = 1) ....
    What events give Y = 2? So what's Pr(Y = 2) ....
    .
    .
    .
    What events give Y = 18? So what's Pr(Y = 18) ....
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  3. #3
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    Here is a bit more.
    Quote Originally Posted by da` View Post
    Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.
    Find the probability function of X and the probability function of Y, if sampling is without replacement.
    Without replacement there are {10 \choose 2}=45 ways to choose two cards.
    That is the denominator of the probabilities.
    Now X can have 9 values: 1,2,,9. X=5 in five ways: {0,5},{1,5},{2,5},{3,5},{4,5}.
    Thus P(X=5)=\frac{5}{45}.
    You do the others.

    Now Y can have values 1,2,,17. Y=8 in four ways: {0,8},{1,7},{2,6},{5,3}.
    Thus P(Y=8)=\frac{4}{45}.
    You do the others.

    Quote Originally Posted by da` View Post
    Two cards are picked from a bag containing 10 cards numbered 0,1,2,.....,9. Let X be the larger number of the two cards and let Y be their total.
    Find the probability function of X and the probability function of Y, if sampling is with replacement.
    With replacement there are {10}^2 =100 ways to choose two cards.
    That is the denominator of the probabilities.
    Now X can have 10 values: 0,1,2,,9. X=5 in six ways: {0,5},{1,5},{2,5},{3,5},{4,5},{5,5}.
    Thus P(X=5)=\frac{6}{100}.
    You do the others.

    Now Y can have values 1,2,,17. Y=8 in five ways: {0,8},{1,7},{2,6},{5,3},{4,4}.
    Thus P(Y=8)=\frac{5}{100}.
    You do the others.
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  4. #4
    da`
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    Thank you very much for your help. My apologies to the moderators for my re-posts.

    Right Ive got the without replacement down fine but its just the with replacement one I am having trouble with.

    Without

    P(X=1) = 1/45
    P(X=2) = 2/45
    P(X=3) = 3/45
    P(X=4) = 4/45
    P(X=5) = 5/45
    P(X=6) = 6/45
    P(X=7) = 7/45
    P(X=8) = 8/45
    P(X=9) = 9/45

    P(Y=1) = 1/45
    P(Y=2) = 1/45
    P(Y=3) = 2/45
    P(Y=4) = 2/45
    P(Y=5) = 3/45
    P(Y=6) = 3/45
    P(Y=7) = 4/45
    P(Y=8) = 4/45
    P(Y=9) = 5/45
    P(Y=10) = 4/45
    P(Y=11) = 4/45
    P(Y=12) = 3/45
    P(Y=13) = 3/45
    P(Y=14) = 2/45
    P(Y=15) = 2/45
    P(Y=16) = 1/45
    P(Y=17) = 1/45

    With (ill explain my problem)

    so 100 ways to choose two cards


    P(X=0) = 1/100 eg {0,0}
    P(X=1) = 2/100 eg {0,1}, {1,1}
    P(X=2) = 3/100 eg {0,2}, {1,2}, {2,2}
    P(X=3) = 4/100
    P(X=4) = 5/100
    P(X=5) = 6/100
    P(X=6) = 7/100
    P(X=7) = 8/100
    P(X=8) = 9/100
    P(X=9) = 10/100

    (1+2+...+10)/100= 55/100

    P(Y=1) = 1/100 eg {0,1}
    P(Y=2) = 2/100 eg {0,2}, {1,1}
    P(Y=3) = 2/100
    P(Y=4) = 3/100
    P(Y=5) = 3/100
    P(Y=6) = 4/100 eg {1,5}, {0,6}, {2,4}, {3,3}
    P(Y=7) = 4/100
    P(Y=8) = 5/100
    P(Y=9) = 5/100
    P(Y=10) = 5/100
    P(Y=11) = 4/100
    P(Y=12) = 4/100
    P(Y=13) = 3/100
    P(Y=14) = 3/100
    P(Y=15) = 2/100
    P(Y=16) = 2/100
    P(Y=17) = 1/100

    (2x1) + (4x2) + (4x3) + (4x4) + (3x5) = 53/100

    which is clearly wrong! what am i doing wrong?
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  5. #5
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    I sorry that mislead you on the replacement case.
    You need to double your answers.
    X=2 happens if (0,2),(2,0),(1,2),(2,1),(2,2) so five ways.
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  6. #6
    da`
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    Ok cool. Thanks.

    Is it true then with replacement Y has values 1,2,...,18. as {9,9} is a possible pick too?
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  7. #7
    da`
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    sorry 0,1,2,...,18 as {0,0} is also a possible pick?
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