There are two urns, each with some numbered cards inside.
In urn A there are cards representing all 3-digit numbers (from 109 to 910 inclusively), having 10 as their sum of digits.
In urn B - all 3-digit numbers (from 119 to 920 inclusively), with 11 as their s.o.d.
Assuming you aim to randomly draw a prime number,
which urn would you choose?
What if there was a 3rd urn with s.o.d=15, would you consider it as a good choice?
(In reply to
re: computer assisted solution by snark)
These are the 42 numbers that account for the denominator in 12/42 ~= .2857 for the first urn:
118 136 145 154 172 190 208 217 226 235 244 253 262 280 316 325 334 343 352 361 370 406 415 424 442 451 460 505 514 532 550 604 622 640 703 712 721 730 802 820 901 910
and the 48 for the 13/48 ~= .27083 for the second urn:
119 128 146 155 164 182 209 218 236 245 254 272 290 308 326 335 344 362 371 380 407 416 425 434 452 470 506 515 524 533 542 551 560 605 614 623 632 650 704 713 722 731 740 803 812 830 902 920
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Posted by Charlie
on 2013-01-10 22:32:28 |