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64-points-are-in-a-plane-x-y-x-0-1-2-7-y-0-1-2-7-4-points-are-chosen-at-random-What-is-the-proabability-the-lines-connecting-them-do-not-form-a-square-or-rectangle-




Question Number 2771 by prakash jain last updated on 26/Nov/15
64 points are in a plane:  (x,y), x∈{0,1,2,...,7}, y∈{0,1,2,...,7}  4 points are chosen at random.  What is the proabability the lines connecting  them do not form a square or rectangle?
64pointsareinaplane:(x,y),x{0,1,2,,7},y{0,1,2,,7}4pointsarechosenatrandom.Whatistheproababilitythelinesconnectingthemdonotformasquareorrectangle?
Answered by Rasheed Soomro last updated on 01/Dec/15
Please Guide me if  I am wrong.  I don′t know much about PROBABILITY.  Assuming that sides of rectangle are  horizantal and vertical.  Assuming that all vertices are different points.  Being P_1 P_2 ^(−)   horizantal, P_1  and P_(2 )  have same y−coordinate.  For same reason P_3  and P_4  have same y−coordinate.  Being P_1 P_4 ^(−)   and P_2 P_3 ^(−)  vertical, P_1  and P_4  have same  x−coordinate, and P_2  and P_(3 ) have same x−coordinate.  Let  P_1 =(x_1 ,y_1 ),P_2 =(x_2 ,y_1 ),P_3 (x_2 ,y_2 )   and   P_4 =(x_1 ,y_2 )  where x_i  ∈{0,1,2,...,7},y_i  ∈{0,1,2,...,7}  We first consider possibility of being rectangle/square._(−)   Possibility of cboosing P_1 ,P_2 ,P_3  and P_4    is Possibility of cboosing their coordinates  x_1 ,y_1 ,x_2   and y_2   Choosing x_1  has (1/8) possibility and y_1  has (1/8) possibility.  Hence choosing P_1  has (1/(64)) possibility.  Choosing x_2  [≠x_1 ] after choosing x_1  has (1/7) possibility.  Choosing x_1 ,y_1  and x_2  has (1/8)×(1/8)×(1/7) =(1/(448)) possibility.  Choosing y_(2 ) after choosing y_1  has (1/7) possibility.   Choosing all the coordinates x_1 ,y_1 ,x_2  and  y_2   in succesion  has (1/8)×(1/8)×(1/7)×(1/7)=(1/(3136)) possibility of being_(−)  rectangle.  Possibility of not being_(−)  rectangle=1−(1/(3136))=((3135)/(3136))
PleaseGuidemeifIamwrong.IdontknowmuchaboutPROBABILITY.Assumingthatsidesofrectanglearehorizantalandvertical.Assumingthatallverticesaredifferentpoints.BeingP1P2horizantal,P1andP2havesameycoordinate.ForsamereasonP3andP4havesameycoordinate.BeingP1P4andP2P3vertical,P1andP4havesamexcoordinate,andP2andP3havesamexcoordinate.LetP1=(x1,y1),P2=(x2,y1),P3(x2,y2)andP4=(x1,y2)wherexi{0,1,2,,7},yi{0,1,2,,7}Wefirstconsiderpossibilityofbeingrectangle/square.PossibilityofcboosingP1,P2,P3andP4isPossibilityofcboosingtheircoordinatesx1,y1,x2andy2Choosingx1has18possibilityandy1has18possibility.HencechoosingP1has164possibility.Choosingx2[x1]afterchoosingx1has17possibility.Choosingx1,y1andx2has18×18×17=1448possibility.Choosingy2afterchoosingy1has17possibility.Choosingallthecoordinatesx1,y1,x2andy2insuccesionhas18×18×17×17=13136possibilityofbeingrectangle.Possibilityofnotbeingrectangle=113136=31353136
Commented by prakash jain last updated on 04/Dec/15
May the question was not clear but the coordiantes  of the points are (x,y) are (0,0)..(0,7)  (1,0) to (1,7)  so you know when two lines are ⊥^r .
Maythequestionwasnotclearbutthecoordiantesofthepointsare(x,y)are(0,0)..(0,7)(1,0)to(1,7)soyouknowwhentwolinesarer.
Commented by Rasheed Soomro last updated on 30/Nov/15
•I understood from  your question: ′the square array  of 64 points of order 8×8 from (0,0) to (7,7)′. Am I wrong?  •Are the sides of rectangle horizantal and vertical only?
Iunderstoodfromyourquestion:thesquarearrayof64pointsoforder8×8from(0,0)to(7,7).AmIwrong?Arethesidesofrectanglehorizantalandverticalonly?
Commented by Rasheed Soomro last updated on 03/Dec/15
What is the meaning of ′ two points are ⊥^r  ′
Whatisthemeaningoftwopointsarer
Commented by prakash jain last updated on 04/Dec/15
⊥^r =perpendicular. Actually I meant lines.  I was travelling for last few days returned  only today.
r=perpendicular.ActuallyImeantlines.Iwastravellingforlastfewdaysreturnedonlytoday.
Commented by Rasheed Soomro last updated on 04/Dec/15
ThAnkS!
ThAnkS!

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