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Math Help - Conditional probability and Expectation

  1. #1
    kin
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    Conditional probability and Expectation


    (ii) i have no ideas.......
    (iii) P(X=3|Y=6) = P(X=3, Y=6) / P(Y=6)
    then ,how to find P(X=3, Y=6)????
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  2. #2
    Flow Master
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    Quote Originally Posted by kin View Post

    (ii) i have no ideas.......
    (iii) P(X=3|Y=6) = P(X=3, Y=6) / P(Y=6)
    then ,how to find P(X=3, Y=6)????
    Y ~ Binomial(n = 10, p = 2/5) and X ~ Binomial(n = 5, p = 2/5).

    (ii) Note that Var(Y) = E(Y^2) - (E(Y))^2 \Rightarrow np(1 - p) = E(Y^2) - (np)^2.

    (iii) Try having another think now.
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  3. #3
    kin
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    Quote Originally Posted by mr fantastic View Post
    Y ~ Binomial(n = 10, p = 2/5) and X ~ Binomial(n = 5, p = 2/5).

    (ii) Note that Var(Y) = E(Y^2) - (E(Y))^2 \Rightarrow np(1 - p) = E(Y^2) - (np)^2.

    (iii) Try having another think now.
    (ii) thanks
    (iii)
    P(X=3,X=6)
    = P(Y=6| X=3) * P(X=3)
    ={ [((3/5)^2) ((2/5)^3) (5C3)]^2 } *[((3/5)^2) ((2/5)^3) (5C3)]
    =[((3/5)^2) ((2/5)^3) (5C3)]^3

    so,
    P(X=3| Y=6)
    = P(X=3,X=6)/ P(Y=6)
    ={[((3/5)^2) ((2/5)^3) (5C3)]^3}/ [((3/5)^4) ((2/5)^6) (10C6)]
    =((3/5)^2) ((2/5)^3)((5C3)^3)/ (10C6)
    =96/875
    AM I RIGHT??
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