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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 10:34:17 IST
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Two small squares on a chess board are chosen at random.What is the probability that they have a common side ?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 11:06:20 IST
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is it 1/18???
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Nitwit Blubber Odment Tweak
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 11:07:22 IST
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2/32=1/16 rate if right
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i am genius |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 11:08:32 IST
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radon pl post the ans..if im rite i can post soln
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Nitwit Blubber Odment Tweak
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 11:19:28 IST
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il pot my soln neway... leave out the outer border of squares... each inner square has 4 neighbours wid common border... now 4 outer border: except the corner sq all have 3 sq wid common border...n the corners have 2 sq wid common border now each pair of sq are mapped together twice so devide by 2.. so v get [{36*4 +24*3+4*2}/2]/64C2=1/18
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Nitwit Blubber Odment Tweak
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 11:47:33 IST
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thanx Gokul
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 12:22:11 IST
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Here's a slightly simpler method .
Count the no. of sides which are common to 2 squares . these include the 7 horizontal and 7 vertical lines which include 8 sides each .
So P= 7x8x2 / 64C2 = 1/18
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 12:34:16 IST
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hey... is the answer 1/64 * 14/63..... here is the logic behind my answer... the first square can be any one of the 64 squares.... for the next square... we hv to choose one of the squares which lie on horizontal or vertical line wrt the 1st square.. there are 14 such square... so the probablity that the nest square is one of those 14 is 14/63
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size doesnt matter, brains do...
vignesh |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 12:54:38 IST
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@ coolvig: ans is 1/18..n u r gettin sumtin else..so..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 12:57:01 IST
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nadeem can u pls explain a bit more...i think ur solution is shortest but not getting it really
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 14:11:36 IST
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We are interested in the 2 square combinations. we want 2 square combinations which have 1 side in common. This is equal to the no. of sides in common. Count the no. of common sides , not the squares.
Each horizontal and vertical line ( except the ones at the end) will be a common side for 2 squares.
Hope u got it.
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