Three fair dice are rolled. What is the probability that the sum of the rolls is a multiple of 4?
4==(112, 121, 211)>>Total = 3 ways
8==(116, 125, 134, 143, 152, 161, 215, 224, 233, 242, 251, 314, 323, 332, 341, 413, 422, 431, 512, 521, 611)>>Total = 21 ways
12==(156, 165, 246, 255, 264, 336, 345, 354, 363, 426, 435, 444, 453, 462, 516, 525, 534, 543, 552, 561, 615, 624, 633, 642, 651)>>Total = 25 ways
16==(466, 556, 565, 646, 655, 664)>>Total = 6 ways
Add up the 4 totals ==3 + 21 + 25 + 6 ==55 ways of getting a multiple of 4
Probability is: 55 / 6^3 ==55 / 216
