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Math Help - About pdf

  1. #1
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    About pdf

    Hello. There are 3 boys, 3 pictures and 3 car names.
    A task is just to connect car names into their pictures.
    If a boy connects car names into pictures randomly, what is a probability function
    on number Y of correct choises?


    Let's see, 3 pictures, 3 cars and because it is enough that one boy connects, so
    that would be: {1,2,3,4,5,6,7} and sum of that is 28.

    ( Unless, we have to deal correct choises: 3 pictures = 1/3. 3 car names = 1/3. ...)


    Next step is division.
    Thus, 1/28, 2/28, 3/28, 4/28, 5/28, 6/28, 7/28.
    2/28 = 1/14, 4/28 = 1/7 and 7/28 = 1/4.
    Hence, pdf ( probability density function ) on number Y of correct choises is
    {1/28, 1/14, 3/28, 1/7, 5/28, 6/28, 1/4}
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  2. #2
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    Re: About pdf

    Quote Originally Posted by TractorGreen View Post
    Hello. There are 3 boys, 3 pictures and 3 car names. A task is just to connect car names into their pictures.
    If a boy connects car names into pictures randomly, what is a probability function on number Y of correct choices?
    Not sure of your reading of this question?
    Surely, the outcomes are \{0,1,2,3,4,5,6,7,8,9\}.
    It could be that all three could be correct: \mathcal{P}(Y=9)=\frac{1}{6^3}.
    Because any boy can get all three correct with probability \frac{1}{6}.
    Have I misunderstood the question?
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  3. #3
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    Re: About pdf

    I am not sure. But maybe you understood correctly my question.
    I'm lookin for pdf ( I mean probability density function ).
    And yes, any boy can connect car names into pictures randomly.
    And any boy can get all three correct.

    That {0,1,2,3,4,5,6,7,8,9}; will that be 3 boys + 3 cars + 3 pictures = 9

    But is that {0,1,2,3,4,5,6,7,8,9} a probability density function on correct choises?
    Because I know that pdf cannot be just a number.
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  4. #4
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    Re: About pdf

    Quote Originally Posted by TractorGreen View Post
    I am not sure. But maybe you understood correctly my question. I'm lookin for pdf ( I mean probability density function ).
    And yes, any boy can connect car names into pictures randomly.
    And any boy can get all three correct.
    That {0,1,2,3,4,5,6,7,8,9}; will that be 3 boys + 3 cars + 3 pictures = 9
    But is that {0,1,2,3,4,5,6,7,8,9} a probability density function on correct choises?
    Because I know that pdf cannot be just a number.
    Each boy can get \{0,1,3\} note it is impossible to get a 2.
    \mathcal{P}(0)=\frac{2}{6},~\mathcal{P}(1)=\frac{3  }{6},~\mathcal{P}(3)=\frac{1}{6}.

    Because no one can get a 2, it is impossible to get 8.

    We can get a 6 by 033,~ 303,~330.

    So you have to figure out all possible sums.
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  5. #5
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    Re: About pdf

    Okay. Here is my suggestion:

    all possible sums for {0,1,3}

    0 = 0, 00, 000
    1 = 1, 01, 10, 001, 010
    3 = 111, 003, 030, 003

    Thus, a probability density function on correct choises is

    {0,00,000,1,01,10,001,010,111,003,030,003}
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  6. #6
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    Re: About pdf

    Quote Originally Posted by TractorGreen View Post
    Okay. Here is my suggestion:
    all possible sums for {0,1,3}
    0 = 0, 00, 000
    1 = 1, 01, 10, 001, 010
    3 = 111, 003, 030, 003
    Thus, a probability density function on correct choises is
    {0,00,000,1,01,10,001,010,111,003,030,003}
    I have no idea what any of that means.
    But \mathcal{P}(Y=6)=3\frac{2}{6}\frac{1}{6}\frac{1}{6  }.
    \mathcal{P}(Y=0)=\frac{2^3}{6^3} because that only happens if 000.

    Now you have to work out how Y=1,2,3,4,5,\&~7.
    Last edited by Plato; October 3rd 2011 at 04:36 AM.
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  7. #7
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    Re: About pdf

    Hello.

    This is difficult, but I sure try.


    0 1 1 1^3
    P(Y = 1) = 1 - - - = ----
    6 6 6 6^3

    1 0 1 1^3
    P(Y = 2) = 1 - - - = ---
    6 6 6 6^3


    1 1 1 1^3
    P(Y = 3) = 1 - - - = ---
    6 6 6 6^3
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  8. #8
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    Re: About pdf

    P(Y = 1) =

    0 0 1 1^2
    1 - - - = ----
    6 6 6 6^3



    P(Y = 2) =

    1 0 1 1^2
    1 - - - = ---
    6 6 6 6^3


    P(Y = 3) =

    1 1 1 1^3
    1 - - - = ---
    6 6 6 6^3


    P(Y = 4) =

    3 1 0 3^1
    2 - - - = ---
    6 6 6 6^3


    P(Y = 5) =


    3 1 1 3^1
    2 - - - = ---
    6 6 6 6^3


    P(Y = 7) =


    3 3 1 3^2
    3 - - - = ---
    6 6 6 6^3


    This is really hard, but I truly tried to think this over. I'm afraid this is still wrong?
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  9. #9
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    Re: About pdf

    Recall this reply:
    Each boy can get \{0,1,3\} note it is impossible to get a 2.
    \mathcal{P}(0)=\frac{2}{6},~\mathcal{P}(1)=\frac{3  }{6},~\mathcal{P}(3)=\frac{1}{6}.
    Note Y=1 in three ways: 100,~010,~001 so  \mathcal{P}(Y=1)= 3\cdot\frac{3\cdot2^2}{6^3} .
    Note Y=2 in three ways: 110,~011,~101 so  \mathcal{P}(Y=2)= 3\cdot\frac{3^2\cdot2}{6^3} .

    Now you must build each of the remaining cases.
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  10. #10
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    Re: About pdf

    This is 3 and 4. I hope these are right.


    Y = 3 in three ways: 111, 003, 300. P(Y = 3) = 3 * 3 * 2^3 / 6^3

    Y = 4 in three ways: 301, 031, 013. P(Y = 4) = 3 * 3^2 * 3 / 6^3
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  11. #11
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    Re: About pdf

    Quote Originally Posted by TractorGreen View Post
    This is 3 and 4. I hope these are right.
    Y = 3 in three ways: 111, 003, 300. P(Y = 3) = 3 * 3 * 2^3 / 6^3
    Y = 4 in three ways: 301, 031, 013. P(Y = 4) = 3 * 3^2 * 3 / 6^3
    You missed one for Y=3: 030.

    You missed three for Y=4: 103,~130,~310.
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  12. #12
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    Re: About pdf

    I can't understand more than this. I hope this is correct stuff;

    Y = 3 in three ways: 111, 003, 300. P(Y = 3) = 3 * 3 * 2^3 / 6^3

    Y = 4 in three ways: 103, 130, 301, 310, 031, 013. P(Y = 4) = 3 * 3^2 * 3 / 6^3

    Y = 5 in three ways: 113,131,311 P(Y = 5) = 3 * 3 * 3 / 6^3

    Y = 6 in three ways: 330, 033, 303 P(Y = 6) = 3 * 3 * 3^2 / 6^3

    Y = 7 in three ways: 331, 133, 313. P(Y = 7) = 3 * 3 * 3^2 / 6^3


    So is that function

    P = {111, 003, 300,103, 130, 301, 310, 031, 013, 113, 131,311,330, 033, 303, 331, 133, 313} ?
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  13. #13
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    Re: About pdf

    I still don't know what is a probability density function on correct choises


    IS IT THESE;

    Y = 3 in three ways: 111, 003, 300. P(Y = 3) = 3 * 3 * 2^3 / 6^3

    Y = 4 in three ways: 103, 130, 301, 310, 031, 013. P(Y = 4) = 3 * 3^2 * 3 / 6^3

    Y = 5 in three ways: 113,131,311 P(Y = 5) = 3 * 3 * 3 / 6^3

    Y = 6 in three ways: 330, 033, 303 P(Y = 6) = 3 * 3 * 3^2 / 6^3

    Y = 7 in three ways: 331, 133, 313. P(Y = 7) = 3 * 3 * 3^2 / 6^3


    OR THIS:

    P = {111, 003, 300,103, 130, 301, 310, 031, 013, 113, 131,311,330, 033, 303, 331, 133, 313} ?


    OR is it like this;

    Y = 3 = {111, 003, 300}
    Y = 4 = {103, 130, 301, 310, 031, 013} and so on. ??
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