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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 11:51:52 IST
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u need to think only abt 1st 5 throws
so it'll be (1/6)^2*(5/6)^3*1/6
the last 1/6 accounts 4 the final throw
but getting 2 6s from 1st 5 is 5C2 ways
so v have 5C2(1/6)^2*(5/6)^3*1/6
EDITED:
seeing tht as usual i posted solution just after akhil did pls dun rate :D
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Nitwit Blubber Odment Tweak
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