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Math Help - probability distribution and expected value

  1. #1
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    probability distribution and expected value

    I have attempted the question, but I am doubting my answer because it is a rational number and the numbers in context are not.

    I am providing the whole question for context:
    Two standard six-sided dice are tossed. Let X be the sum of the scores on the two dice.
    (a) Find
    (i) P(X=6)
    (ii) P (X>6)
    (iii) P(X=7 | X>5)

    (b) Elena plays a game where she tosses two dice.
    If the sum is 6, she wins 3 points.
    If the sum is greater than 6, she wins 1 point.
    If the sum is less than 6, she loses kpoints.

    Find the value of k for which Elena's expected number of points is zero.


    I need verification with part b).
    What I did was construct a probability distribution table where X was k, 3, and 1. And my probabilities were 10/36, 5/36, 21/36, respectively. Then I set the expected value equal to zero:
    (10k/36) + (3 x 5/36) + (1 x 21/36) =0
    to solve for k.

    I get k as -36/10.

    But I'm not sure if this is right, can someone please check if my method is correct?
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  2. #2
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    Quote Originally Posted by bhuang View Post

    I am providing the whole question for context:
    Two standard six-sided dice are tossed. Let X be the sum of the scores on the two dice.
    (a) Find
    (i) P(X=6)
    (ii) P (X>6)
    (iii) P(X=7 | X>5)

    You should make a two row table. first row x= 2,3\dots ,12 with a second row being P(x) there will be some symmetry here.
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  3. #3
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    My probability distribution table is my two row table. Instead of 2, 3, ... I put the points obtained from the game in. It should be the same thing, right? What do you mean by symmetrical...
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  4. #4
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    Quote Originally Posted by bhuang View Post
    My probability distribution table is my two row table. Instead of 2, 3, ... I put the points obtained from the game in. It should be the same thing, right? What do you mean by symmetrical...
    The sum of two dice can only be from 2 to 12. These are the x's.

    You then need to find the probabililty of each of these occuring.

    I.e P(2) =\frac{1}{36},  P(3) =\frac{2}{36}, P(4) =\frac{3}{36}, \dots ,P(12) =\frac{1}{36}

    Do you know how I found these? After you have this information you can answer the questions.
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