Respuesta :
Answer:
The expected payoff for this game is -$1.22.
Step-by-step explanation:
It is given that a pair of honest dice is rolled.
Possible outcomes for a dice = 1,2,3,4,5,6
Two dices are rolled then the total number of outcomes = 6 × 6 = 36.
[tex]\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}[/tex]
The possible ways of getting a total of 7,
{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }
Number of favorable outcomes = 7
Formula for probability:
[tex]Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
So, the possibility of getting a total of 7 = [tex]\frac{6}{36}=\frac{1}{6}[/tex]
The possible ways of getting a total of 11,
{(5,6), (6,5)}
So, the probability of getting a total of 11 = [tex]\frac{2}{36}[/tex] = [tex]\frac{1}{18}[/tex]
Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,
So, the probability of getting the sum of numbers other than 7 or 11 = [tex]\frac{28}{36}[/tex] = [tex]\frac{7}{9}[/tex]
Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,
Hence, the expected value for this game is
[tex]\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11[/tex]
[tex]\frac{11}{3}+\frac{11}{3}-\frac{77}{9}[/tex]
[tex]\frac{22}{3}-\frac{77}{9}[/tex]
[tex]\frac{66-77}{9}[/tex]
[tex]-\frac{11}{9}[/tex]
[tex]-1.22[/tex]
Therefore the expected payoff for this game is -$1.22.
