Understanding Probability Functions

[Beowulf2007]

Homework Statement

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

Deduce the following three probability functions.

Probability that the number of eyes are red

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

Probability that the number of eyes are either red or white

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

Probability that the number of eyes are either red and white

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

The Attempt at a Solution

My Proof (1):

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

\frac{6}{36} = \frac{1}{6} = P_{R}(t)


My Proof(2):

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

<br /> \begin{array}{|c| c| c| c| c| c| } <br /> \hline<br /> (1,1) &amp; (1,2) &amp; (1,3) &amp; (1,4) &amp; (1,5) &amp; (1,6)\\<br /> \hline<br /> (2,1) &amp; (2,2) &amp; (2,3) &amp; (2,4) &amp; (2,5) &amp; (2,6)\\<br /> \hline<br /> (3,1) &amp; (3,2) &amp; (3,3) &amp; (3,4) &amp; (3,5) &amp; (3,6)\\<br /> \hline<br /> (4,1) &amp; (4,2) &amp; (4,3) &amp; (4,4) &amp; (4,5) &amp; (4,6)\\<br /> \hline<br /> (5,1) &amp; (5,2) &amp; (5,3) &amp; (5,4) &amp; (5,5) &amp; (5,6)\\<br /> \hline<br /> (6,1) &amp; (6,2) &amp; (5,3) &amp; (5,4) &amp; (6,5) &amp; (6,6)\\<br /> \hline<br /> \end{array}<br />
Thus by in the schema:

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

which can be describe by the function:

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

Proof(3)

Thus by in the schema:

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


which can be describe by the function:

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

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

SIncerely Yours
Beowulf

 

[HallsofIvy]

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?

 

[Beowulf2007]

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:

S = \{r,w\}|r,w = 1,2,\ldots, 6\}

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) = r \wee w (lowest number of eyes)

Z(r,w) = r \land w (largest number of eyes).

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

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

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

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

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.