thefibonacci wrote:
anhlukas wrote:
What is the probability of getting a sum of 12 when rolling 3 dice simultaneously?
A.10/216
B.12/216
C.21/216
D.23/216
E.25/216
A.10/216
B.12/216
C.21/216
D.23/216
E.25/216
let the dices be a,b,c
a+b+c = 12
but 0 < a,b,c < 6
so we can re-write the equation as
(6-a)+(6-b)+(6-c) = 12
=> a+b+c = 6
which has total of C(8,2) whole number solutions = 28
but out of these 28, three cases must be where 2 of a,b, or c is 0 and the other is 6. so we need to remove those cases which would be 3 cases ( C(3,2)
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