Three fair dice are rolled. What is the probability that the sum of the rolls is a multiple of 4?
There are 55 combinations which are all multiples of 4 as follows:
112 , 116 , 121 , 125 , 134 , 143 , 152 , 156 , 161 , 165 , 211 , 215 , 224 , 233 , 242 , 246 , 251 , 255 , 264 , 314 , 323 , 332 , 336 , 341 , 345 , 354 , 363 , 413 , 422 , 426 , 431 , 435 , 444 , 453 , 462 , 466 , 512 , 516 , 521 , 525 , 534 , 543 , 552 , 556 , 561 , 565 , 611 , 615 , 624 , 633 , 642 , 646 , 651 , 655 , 664 , Total = 55
Therefore, the probability is= 55 / 6^3 =55 / 216
