11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
-------------------------
Solution to the given sum on probability:
------------------------------------------------------
The total number of ways in which any number can appear
in the three dice = 6x6x6 = 216
The number of ways in which the same number can appear on each
of the three dice =6
(111,222,333,444,555,666)
Therefore the required probability = 6/216 = 1/36.
----------------------------------------------------------------------
For everyone to understand the sum, the entire sample space
containing the total number of cases i.e 216 ways is given below.
From this, anyone can easily identify the number of favourable
cases and hence the required probability.
111 112 113 114 115 116
121 122 123 124 125 126
131 132 133 134 135 136
141 142 143 144 145 146
151 152 153 154 155 156
161 162 163 164 165 166
211 212 213 214 215 216
221 222 223 224 225 226
231 232 233 234 235 236
241 242 243 244 245 246
251 252 253 254 255 256
261 262 263 264 265 266
311 312 313 314 315 316
321 322 323 324 325 326
331 332 333 334 335 336
341 342 343 344 345 346
351 352 353 354 355 356
361 362 363 364 365 366
411 412 413 414 415 416
421 422 423 424 425 426
431 432 433 434 435 436
441 442 443 444 445 446
451 452 453 454 455 456
461 462 463 464 465 466
511 512 513 514 515 516
521 522 523 524 525 526
531 532 533 534 535 536
541 542 543 544 545 546
551 552 553 554 555 556
561 562 563 564 565 566
611 612 613 614 615 616
621 622 623 624 625 626
631 632 633 634 635 636
641 642 643 644 645 646
651 652 653 654 655 656
661 662 663 664 665 666
--------------------------------------------------------------------------
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
-------------------------
Solution to the given sum on probability:
------------------------------------------------------
The total number of ways in which any number can appear
in the three dice = 6x6x6 = 216
The number of ways in which the same number can appear on each
of the three dice =6
(111,222,333,444,555,666)
Therefore the required probability = 6/216 = 1/36.
----------------------------------------------------------------------
For everyone to understand the sum, the entire sample space
containing the total number of cases i.e 216 ways is given below.
From this, anyone can easily identify the number of favourable
cases and hence the required probability.
111 112 113 114 115 116
121 122 123 124 125 126
131 132 133 134 135 136
141 142 143 144 145 146
151 152 153 154 155 156
161 162 163 164 165 166
211 212 213 214 215 216
221 222 223 224 225 226
231 232 233 234 235 236
241 242 243 244 245 246
251 252 253 254 255 256
261 262 263 264 265 266
311 312 313 314 315 316
321 322 323 324 325 326
331 332 333 334 335 336
341 342 343 344 345 346
351 352 353 354 355 356
361 362 363 364 365 366
411 412 413 414 415 416
421 422 423 424 425 426
431 432 433 434 435 436
441 442 443 444 445 446
451 452 453 454 455 456
461 462 463 464 465 466
511 512 513 514 515 516
521 522 523 524 525 526
531 532 533 534 535 536
541 542 543 544 545 546
551 552 553 554 555 556
561 562 563 564 565 566
611 612 613 614 615 616
621 622 623 624 625 626
631 632 633 634 635 636
641 642 643 644 645 646
651 652 653 654 655 656
661 662 663 664 665 666
--------------------------------------------------------------------------
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