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Math Help - Joint probability question???

  1. #1
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    Oct 2008
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    Question Joint probability question???

    hi i have a problem with the last part!! could some1 help me out.

    Two players, A and B, play a rolling dice game. The player whose turn it is chooses wheather to roll one or two dice. if the total showing on the dice is even, the player who rolled them wins that amount of money from the other player, but if the toatal is odd he pays that amount to the other player.

    i) Player A always rolls one die. Let Y denote the score on the die, X the net amound Player A wins, so that X=Y if is even, X=-Y if Y is odd.
    a)Write down in tabular from the joint probabilty function of x and y.

    Player B allways rolls 2 dice. Let the random variable U denote player B`s net gain when he rolls the dice
    iia)Calculate the mean and variance of U
    iib)In ten sucessive rolls of the dice by player B, What is the probability that the player gets at least seven even scores?

    iii)The player rolls alternatly,each making 10 rolls.At the End of that time Player A calculates his net gain G=X(1)-U(1)+X(2)-U(2) +...+ X(10)-U(10). Write down the expectation and variance of G and hence calculate ({apporximatly)) the prob that player A loses money on the game.

    i have these answers for
    7ia) E(x)=0.5 and Var(x)=14.9
    7ic) E(xy)=3.5 hence Cov(xy)=0.265
    7iia) E(u)=0 and E(u^2)= 54.83333... hence VAR(U)=54.833333
    7iib) P(even) = 0.5 hence p(odd)=0.5
    using binomial expansion from 10C7 to 10C8, adding all answers i obtained for the probabilty for at least seven even scores to be 0.17188

    7iii) for this i was very stuck!!! using the rules for linear combinations of norm distributed variables i obtained for the
    E(G)=10E(x)-10E(U)
    E(G)=(10 x 0.5) -(10 x 0)
    E(G)=5-0 hence E(G)=0

    For the var(10(X1) - 10(u1))
    =(10^2 x Var(x)) + (10^2 x var(u))
    =(100 * 14.9) + (100 x 54.833)
    =1490 + 5483.3333
    =6973.3

    but after this for the hence part i have no idea how no incoperate these values in to finding a probability. I think my values for e(g) are correct.
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  2. #2
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    Oct 2008
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    Question

    umm my variance value may be incorrect, ive used the rule
    =Var(x1 - Y2 + X2 -Y2 .......)
    =(A^2 * VAR(X) )+( B^2 * VAR(Y) )
    where A and B are the number of time x-y apperar in the sequencce?
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