Three dice are rolled. Find the probability that all the rolls are different, and none of them are equal to six.
+1005 6 outcomes/ dice, so 6^3 = 216 total outcomes
There are 6 ways of getting a trio of rolls with same numbers, 111,222,...666
We also have the ones with 2 numbers the same. 112,113, etc.
Each of the repeating digit(there's 6) has 3 positions to put the other non-same number, front, center, last, and 5 numbers to choose from, so 6*5*3 = 90 ways.
We ALSO have the ones with 3 different rolls but equal to 6. Each of these has 3! = 6 ways of arrangements.
1+2+3
I honestly don't think there are any else.
SO that's 6.
Adding all the ways, 6+6+90 = 102
So I'm pretty sure the answer's 102/216 = 51/108 and you can simplify that
If you don't understand feel free to ask.
When you say "None of them is equal to six", do you mean the sum of each permutation, or 6 itself must not appear in any permutation?
If that is what he/she meant, then I get a different result as follows:
124 , 125 , 126 , 134 , 135 , 136 , 142 , 143 , 145 , 146 , 152 , 153 , 154 , 156 , 162 , 163 , 164 , 165 , 214 , 215 , 216 , 234 , 235 , 236 , 241 , 243 , 245 , 246 , 251 , 253 , 254 , 256 , 261 , 263 , 264 , 265 , 314 , 315 , 316 , 324 , 325 , 326 , 341 , 342 , 345 , 346 , 351 , 352 , 354 , 356 , 361 , 362 , 364 , 365 , 412 , 413 , 415 , 416 , 421 , 423 , 425 , 426 , 431 , 432 , 435 , 436 , 451 , 452 , 453 , 456 , 461 , 462 , 463 , 465 , 512 , 513 , 514 , 516 , 521 , 523 , 524 , 526 , 531 , 532 , 534 , 536 , 541 , 542 , 543 , 546 , 561 , 562 , 563 , 564 , 612 , 613 , 614 , 615 , 621 , 623 , 624 , 625 , 631 , 632 , 634 , 635 , 641 , 642 , 643 , 645 , 651 , 652 , 653 , 654 , Total = 114
So, the probability is: 114 / 6^3 =19 /36
+1005 No... I might not have been clear.
what I meant to say was that it could be that the three rolls have a total sum of 6, and your answers include sums of more than 6.
