Three different squares are chosen randomly on a chessboard.
What is the probability that they lie in the same diagonal?
(In reply to
re: Corrected solution by SilverKnight)
Touche !! My first two terms should have read
(4/64)*(7/63)*(6/62) and (8/64)*(6/63)*(5/62). When these two corrections are made, I get exactly the same answer you do, so this must be right. For instance, my third line,
(8/64)*(2/63)*(1/62) + (8/64)*(5/63)*(4/62),
reflects the fact that the first randomly selected square has an 8/64 chance to be in the set
{c1, f1, a3, h3, a6, h6, c8, f8}. Each of these is on one diaganol with 2 other squares, and one with 5 other squares.
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(I think Al Gore's invention of the "Al-Gore-ithm" is right up there with George Bush's originating the Code of Bushido).
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Posted by Penny
on 2004-01-14 19:59:55 |
re(2): Corrected solution
