We have a six-sided die
possible outcome
[tex]1,\text{ 2,3,4,5,6}[/tex]
if you roll a 6 you get $13
if you roll a 4 or 5 you get $4
if otherwise, you get 1,2,3 you will them $6
[tex]\begin{gathered} \text{Probability of getting \$13=}\frac{1}{6} \\ \\ \text{Probability of getting \$4=}\frac{2}{6}=\frac{1}{3} \\ \\ \text{Probability of paying them \$6=}\frac{3}{6}=\frac{1}{2} \end{gathered}[/tex]
We are asked to complete the table from smallest to biggest
Part B
The expected profit.
[tex]\begin{gathered} -6(\frac{1}{2})+4(\frac{1}{3})+13(\frac{1}{6}) \\ \\ =-3+\frac{4}{3}+\frac{13}{6} \\ =\text{ \$0.50} \end{gathered}[/tex]
Part C
If you play many games you will likely win on average very close to $0.50 per game because the expected profit is $0.50
Part D
Yes, since the expected value is positive you will likely to come home with more money if you played more games
