Explanation:
The question involves getting the expected value or the sum in a deal involving dice
First, we will have to get the probabilities
Probability is defined by
[tex]P=\frac{number\text{ of possible outcomes}}{number\text{ of total outcome}}[/tex]
If a six is rolled, the probability will be
[tex]P(6)=\frac{1}{6}[/tex]
When a 4 or 5 is rolled, the probability will be
[tex]\begin{gathered} P(4\text{ or 5})=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} \\ P(4\text{ or 5})=\frac{1}{3} \end{gathered}[/tex]
When a 1, 2, or 3 is rolled
[tex]\begin{gathered} P(1\text{ or 2 or 3})=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2} \\ P(1\text{ or 2 or 3})=\frac{1}{2} \end{gathered}[/tex]
We can now get the expected pay using the relationship
[tex]Expected\text{ pay =x.P\lparen x\rparen}[/tex]
So for rolling a 6, the expected will be
[tex]x.P(x)=\text{ \$}12\times\frac{1}{6}=\text{ \$2}[/tex]
When a 4 or 5 is rolled
[tex]x.P(x)=\text{ \$}6\times\frac{1}{3}=\text{ \$2}[/tex]
When 1,2, or 3 are rolled
[tex]x.P(x)=-\text{ \$}5\times\frac{1}{2}=-\text{ \$2.5}[/tex]
We can now construct the table as follow
