I don't think so that probability would be exact 1/1000 . suppose number is ( a9 a8 a7 a6 a5 a4 a3 a2 a1 a0) where a0 is least and a9 is most significant bit then you can generate each of the bit ai using given bit generator
but if but at a time (a9 , a8 , a7 , a6 , a5 , a3 ) and any other bit are set together then number would exceed given limit 1000 which we don't want . so dave's solution generate numbers with probability 1/(2^n) where n is ceil(log2(b-a+1)) so 1/2^10 = 1/1024 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
