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 Two urns (Posted on 2013-01-10)
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?

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): computer assisted solution | Comment 3 of 5 |
(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

 Posted by Charlie on 2013-01-10 22:32:28
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