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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

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 re(4): computer assisted solution -remarki Comment 5 of 5 |

Indeed there was a bug:extra NEXT's in the IF statements.

Corrected:

5   Digsum=10
10      for N=100 to 999
20      S=fnSod(N)
30        if S=Digsum and prmdiv(N)=N then inc Ct
35        if S=Digsum then print N;:inc Ct2
40      next
50      print:print Ct,Ct2,Ct/Ct2
60      end
1070     fnSOD(X)
1080        Sod=0
1090        S=cutspc(str(X))
1100        for I=1 to len(S)
1110          Sod=Sod+val(mid(S,I,1))
1120        next
1130      return(Sod)
OK
run
109  118  127  136  145  154  163  172  181  190  208  217  226  235  244  253 262  271  280  307  316  325  334  343  352  361  370  406  415  424  433  442 451  460  505  514  523  532  541  550  604  613  622  631  640  703  712  721 730  802  811  820  901  910

new stats:
12  out of  54 for prob   0.2222222222222222221

and for sod=11:

119  128  137  146  155  164  173  182  191  209  218  227  236  245  254  263 272  281  290  308  317  326  335  344  353  362  371  380  407  416  425  434 443  452  461  470  506  515  524  533  542  551  560  605  614  623  632  641 650  704  713  722  731  740  803  812  821  830  902  911  920

for stats:
13 out of  61 making prob =   0.2131147540983606557

These probabilities agree with snark's reported results.

Same choice of urn for best chances, as noted by snark.

 Posted by Charlie on 2013-01-11 01:57:35

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