Here is the complete question.
1). What are the odds for rolling a sum of 5 in a single roll of two fair dice?
2). If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?
Answer:
1). [tex]\frac{1}{8}[/tex]
2). $8.
Step-by-step explanation:
1).
The sample space (S) for rolling two fair dice is given as the following parameters illustrated below:
[tex]\left[\begin{array}{cccccc}(1,2)&(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)\end{array}\right] \left[\begin{array}{cccccc}(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)\end{array}\right][/tex]
Let F represent the event that the sum of both dice turns up is 5.
F = [(1,4),(2,3),(3,2),(4,1)}
∴ The probability for an event F is illustrated below as:
[tex]\frac{P(F)}{P(F')}[/tex] [tex]=\frac{\frac{4}{36} }{1-\frac{4}{36} }[/tex]
[tex]=\frac{\frac{4}{36} }{\frac{32}{36} }[/tex]
[tex]={\frac{4}{36}}*\frac{36}{32}[/tex]
= [tex]\frac{1}{8}[/tex]
2). From above, we can see that the probability for rolling a sum of 5 are 1 to 8. Therefore, if you roll a sum of 5, in order for the game to be fair, the house is required to pay $8.
