SOLUTION: I'm having issues trying to get this problem down pact, I dont know what im doing wrong. Craps. In the game of craps, the player rolls two balanced dice. Thirty-six equally l

Question 1087493: I'm having issues trying to get this problem down pact, I dont know what im doing wrong.

Craps. In the game of craps, the player rolls two balanced dice. Thirty-six equally likely
outcomes are possible, as shown
A = event the sum of dice is 6
B = event the sum of dice is 10
C = event the sum of dice is 3
D = event the sum of dice is 5
E = event the sum of dice is 9
F = event the sum of dice is 4 and
a. Compute the probability of each of the six events.
b. Compute the probability the player wins on the first roll if the sum of the dice is a 5 or 9
c. Compute the probability the player loses on the first roll if the sum of the dice is a 3, 4
or 6

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The probability of a 7 is 6/36
For every number on each side it decreases 1/36, so a 6 or 8 would be 5/36, a 5 or 9, 4/36, 4 and 10 3/36, and 3 and 11, 2/36, and 2 and 12, 1/36
That will answer the first 6 questions.
5 or 9 will be 8/36
loses will have a probability of 10/36.
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