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Math Help - Dice

  1. #1
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    Dice

    Hi, I am currently stuck on this extended response question from my book, I really don't know how to do the parts of the questions, could anyone help me? Thanks

    You are given two un-biased dice, one painted red and one painted blue. You roll the dice and you record the numbers showing on each die. Let R by the random variable for the red die and B the corresponding random variable for the blue die.

    (a) Write down the probability distribution for each of the random variables R and B.
    (b) Write down the joint probability distribution, Pr(R,B). That is, the probability of obtaining, in the one roll, a speci c value for R and a speci c value for B.
    (c) Use the joint distribution to deduce the probability distribution for R - B.
    (d) Hence calculate the mean and standard deviation for R - B.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by usagi_killer View Post
    Hi, I am currently stuck on this extended response question from my book, I really don't know how to do the parts of the questions, could anyone help me? Thanks

    You are given two un-biased dice, one painted red and one painted blue. You roll the dice and you record the numbers showing on each die. Let R by the random variable for the red die and B the corresponding random variable for the blue die.

    (a) Write down the probability distribution for each of the random variables R and B.
    (b) Write down the joint probability distribution, Pr(R,B). That is, the probability of obtaining, in the one roll, a speci c value for R and a speci c value for B.
    (c) Use the joint distribution to deduce the probability distribution for R - B.
    (d) Hence calculate the mean and standard deviation for R - B.
    So what have you done? You should be able to do (a) and (b) straight off.

    When you have done (b) for each point (r,b) in the sample space you have a probability, and you can compute r-b, then ...

    CB
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  3. #3
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    Post

    a) P(R) = P(B) = \frac{1}{6}

    b) P(R,B) = \frac{1}{36}

    c) Y= R-B

    ran(Y) = \{-5,-4,-3,-2,-1,0,1,2,3,4,5\}

    P(-5) = P(5) = \frac{1}{36}

    P(-4) = P(4) = \frac{2}{36}

    P(-3) = P(3) = \frac{3}{36}

    P(-2) = P(2) = \frac{4}{36}

    P(-1) = P(1) = \frac{5}{36}

    P(0) = \frac{6}{36}

    d) Positive and negative factors are canceling, so \mu(Y)=E(Y)=0

    \sigma = \sqrt{E(Y^2) - \mu^2} = \sqrt{(\frac{25}{36} +\frac{16}{36} +\frac{9}{36} +\frac{4}{36} + \frac{1}{36}) \times 2 - 0} = \sqrt{3.0555} \approx 1.74
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