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Homework Help: Throwing Dice

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Given [tex]S = \{(r,w)| r,w = 1,2,\ldots, 6\}[/tex]

    Deduce the following three probability functions.

    Probability that the number of eyes are red

    (1)[tex]P_{R}(t) = \frac{1}{6}[/tex] for [tex]t \in \{1,2,\ldots 6 \}[/tex]

    Probability that the number of eyes are either red or white

    (2)[tex]P_{Y}(t) = \frac{13-2t}{36}[/tex] for [tex]t \in \{1,2,\ldots 6 \}[/tex]

    Probability that the number of eyes are either red and white

    (3)[tex]P_{Z}(t) = \frac{2t-1}{36}[/tex] for [tex]t \in \{1,2,\ldots 6 \}[/tex]


    3. The attempt at a solution

    My Proof (1):

    Since there is 6 sides on each dice the combined space [tex]S = 6 \cdot 6 = 36 [/tex] and since there is 6 sides on each sides of red dice, then

    [tex]\frac{6}{36} = \frac{1}{6} = P_{R}(t)[/tex]


    My Proof(2):

    The Events of throwing the two dice are describe in the schema:

    [tex]
    \begin{array}{|c| c| c| c| c| c| }
    \hline
    (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\
    \hline
    (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\
    \hline
    (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\
    \hline
    (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\
    \hline
    (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\
    \hline
    (6,1) & (6,2) & (5,3) & (5,4) & (6,5) & (6,6)\\
    \hline
    \end{array}
    [/tex]
    Thus by in the schema:

    [tex]\begin{array}{ccc} P(x = 1) = \frac{11}{36} & P(x = 2) = \frac{9}{36} & P(x = 3) = \frac{7}{36}\\P(x = 4) = \frac{5}{36} & P(x = 5) = \frac{3}{36} & P(x = 6) = \frac{1}{36} \end{array}[/tex]

    which can be describe by the function:

    [tex]P_{Y}(t) = \frac{13-2t}{36}[/tex] for [tex]t \in \{1,2,\ldots 6 \}[/tex]

    Proof(3)

    Thus by in the schema:

    [tex]\begin{array}{ccc} P(x = 1) = \frac{1}{36} & P(x = 2) = \frac{3}{36} & P(x = 3) = \frac{5}{36}\\P(x = 4) = \frac{7}{36} & P(x = 5) = \frac{9}{36} & P(x = 6) = \frac{11}{36} \end{array}[/tex]


    which can be describe by the function:

    [tex]P_{Y}(t) = \frac{2t-1}{36}[/tex] for [tex]t \in \{1,2,\ldots 6 \}[/tex]

    What You Guys say I have deduced the probability functions correctly?? Am I on the right track??

    SIncerely Yours
    Beowulf
     
    Last edited: Nov 26, 2007
  2. jcsd
  3. Nov 27, 2007 #2

    HallsofIvy

    User Avatar
    Science Advisor

    I am totally baffled by this! What in the world does the probability of a die coming up a particular number, or anything you have done, have to do with "eyes" being red or white?
    Your definition of S says nothing about "eyes" and there is no mention of "eyes" in anything except the questions! What "eyes" are you talking about?
     
  4. Nov 27, 2007 #3
    Hello Hallsoft:

    Thank You for Your reply. I am sorry that the context of the problem was not clear.

    Here is the problem in its full context:

    Suppose we throw a red and white die simultaneously. The possible outcome of an experiment such as this can be recorded as follows:

    [tex]S = \{r,w\}|r,w = 1,2,\ldots, 6\}[/tex]

    where r is the number of eyes the die shows and w is the number of eyes that the white die shows.

    R, Y, Z are random variables on the Space S and are defined as follows:

    R(r,w) = r (number of eyes that the red dice shows)

    Y(r,w) = [tex]r \wee w[/tex] (lowest number of eyes)

    Z(r,w) = [tex]r \land w[/tex] (largest number of eyes).

    What I am tasked with showing is that these random varibles can be expressed by probability functions below:

    [tex]p_{R} (t) = p_{W}(t) = \frac{1}{6}[/tex] where [tex]t \in \{1,2,\ldots,6\}[/tex]

    [tex]p_{Y} (t) = \frac{13-2t}{36}[/tex] where [tex]t \in \{1,2,\ldots,6\}[/tex]

    [tex]p_{Y} (t) = \frac{2t-1}{36}[/tex] where [tex]t \in \{1,2,\ldots,6\}[/tex]

    This was what I am was trying to show in my orginal post. Does the assigment justify my atempted solution?

    Thank You for Your answer in Advance.

    Sincerely Yours
    Beowulf.
     
    Last edited: Nov 27, 2007
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