Sajjad1994 wrote:
In the popular casino game of craps, the pass line bet wins when a pair of 6-sided dice is rolled and the sum is a seven or eleven. The pass line bet loses when the sum is two, three or twelve. Any other number from two to twelve that are not one of the aforementioned numbers becomes "the point" and no money is won or lost on that roll.
From the table below, select the expression that represents the probability of the pass line bet losing on the first roll and the probability of neither winning nor losing on the first roll.
Losing | Neither | |
| | 1/12 |
| | 1/9 |
| | 3/10 |
| | 11/18 |
| | 2/3 |
| | 13/18 |
A pair of 6 sided dice are rolled
Total number of outcomes = 6*6 = 36
Winning outcomes: Sum = 7/11
Losing outcomes: Sum = 2/3/12
Neither outcomes: Sum = 4/5/6/8/9/10
Let us examine the first case:
Probability (Losing on the first roll)Losing outcomes: Sum = 2/3/12
Sum = 2: Possible outcomes = (1,1)
Sum = 3: Possible outcomes = (1,2) (2,1)
Sum = 12: Possible outcomes = (6,6)
Probability = Number of possible outcomes / Total number of outcomes = 4/36 = 1/9Let us examine the second case:
Probability (Neither winning nor losing on the first roll)Neither outcomes: Sum = 4/5/6/8/9/10
Sum = 4: Possible outcomes = (1,3) (3,1) (2,2)
Sum = 5: Possible outcomes = (1,4) (4,1) (2,3) (3,2)
Sum = 6: Possible outcomes = (1,5) (5,1) (2,4) (4,2) (3,3)
Sum = 8: Possible outcomes = (2,6) (6,2) (3,5) (5,3) (4,4)
Sum = 9: Possible outcomes = (3,6) (6,3) (4,5) (5,4)
Sum = 10: Possible outcomes = (4,6) (6,4) (5,5)
Probability = Number of possible outcomes / Total number of outcomes = 24/36 = 2/3Losing: 1/9 ; Neither: 2/3