a 4*4 matrix has 16 points. find the probability of making triangle from this points?

• 123/140
  16c3-(10*4c3+4*3c3)
  triangle will not form when points are in straight line
  516
  516/540=123/140
• 9 years agoHelpfull: Yes(18) No(7)
• ans= 516/560
  
  
• 9 years agoHelpfull: Yes(14) No(1)
• Ans will be 516/560 i.e 129/140
• 9 years agoHelpfull: Yes(7) No(0)
• 4 x 4 matrix:- . . . .
  . . . .
  . . . .
  . . . .
  
  so, total possible triangles formed = 16c3
  and,
  for 4 rows and 4 columns :- 4c3 * (4 +4) = 32
  for 2 diagonals formed:- 4c3 * 2 = 8
  for side-diagonals with 3 points:- 3c3 * 4 = 4
  
  So, probability of forming a triangle,
  1 - (32 + 8 + 4)/16c3 = 516/560
• 7 years agoHelpfull: Yes(5) No(0)
• ans: 67/70
  
  Sol: we have to choose 3 points out of 16 point.
  So, 16C3
  now among them 4points in 4 col and 4 row..... we cannot choose 3 points from them so
  16C3-(4C3*8)
  now we can take 4 points to draw one side of the triangle...... so it is 8C1
  So total triangle we can make 16C3-(4C3*8) +8C1= 536
  
  So, the probability of Triangle is 536/16C3
  = 536/560
  =67/70
  
  
  Pls correct me If I am wrong
• 9 years agoHelpfull: Yes(1) No(12)
• This was asked in elitmus exam on 5th july ? results are out ?
  
• 9 years agoHelpfull: Yes(0) No(0)
• The probability of amkeing triangle from 16 points is 28
• 9 years agoHelpfull: Yes(0) No(8)
• 4*(4c1*4c2)/16c3
  
• 9 years agoHelpfull: Yes(0) No(2)
• if there are no co-linear points answer would be 16c3
  but there are co-linear points
  there are 9 straight lines which each contain 4 points
  2 straight lines which each contain 3 points.
  we have to exclude these points
  so the ans=16c3 - (9 X 4c3) - (2 X 3c3)
  =102
  
• 9 years agoHelpfull: Yes(0) No(1)