Throwing Dice

Beowulf2007

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

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:

[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.