Two squares are chosen on a chessboard at random. What is the probability that they have a side in common?
a) 1/18
b) 64/4032
c) 63/64
d) 1/9

• Ans: a) 1/18
  
  In 64 squares, there are:
  
  (1) 4 at-corner squares, each has ONLY 2 squares each having a side in common with...
  (2) 6*4 = 24 side squares, each has ONLY 3 squares such that each has a side in common with...
  (3) 6*6 = 36 inner squares, each has 4 squares such that each has a side in common with...
  So we have the calculation:
  
  P = (4/64)*(2/63) + (24/64)*(3/63)+ (36/64)*(4/63)
  P = 1/18
  
• 8 years agoHelpfull: Yes(20) No(3)
• Solution:
  
  Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.
  So sample space =64C2
  
  Now there are 7 unique adjacent square sets in each row and each column.
  i.e. favourable cases will be 7×(8 rows + 8 columns) = 112.
  Hence Required Probability=(
  Favourable cases
  Sample Space
  )
  =
  112
  64C2
  
  = 1/18
• 8 years agoHelpfull: Yes(13) No(2)
• Ana is 1/9
  Solution:
  4 squares are corner ones which will have 2 adjacent squares each. So, no of ways of choosing corner square and an adjacent square are 4C1*2C1
  Similarly, ways of choosing an edge square and it's adjacent square are 24C1*3C1
  
  Remaining squares=64-4-24=36
  Ways of choosing middle square and it's adjacent squares are 36C1*4C1
  
  Total no of ways of choosing any two squares is 64C2
  
  Therefore probability= (8+72+144)/64C2
  =1/9
• 8 years agoHelpfull: Yes(1) No(4)
• Answer is 1/9.
  4 squares at corner share 2 squares. So 4*2=8
  Remaining 24 sidemost squares share 3 squares 24*3=72
  Remaining 36 squares share 4 squares so 36*4=144
  total no. of ways = 8+72+144=224
  total no of ways selecting 2 from 64 squares is 64C2
  so probability=224/64c2=(1/9)
  
• 8 years agoHelpfull: Yes(1) No(4)
• I think its 2/9
  sample space is 64c2
  consider single row having 8 squares let number'em as 1,2,3..8
  for a single row we have selecting two squares is 8*7 ways
  but,(1,2)(2,1) are same bcz order is not important so total ways is (8*7)/2=>28
  for 8 rows we have 28*8 selections
  similarily
  for columns we have 28*8 selections
  so total=28*8+28*8=>448
  probability is 448/64c2=>2/9
• 8 years agoHelpfull: Yes(0) No(5)
• Ans-(a)1/18
• 8 years agoHelpfull: Yes(0) No(1)
• Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.
  So sample space =64C2
  
  Now there are 7 unique adjacent square sets in each row and each column.
  i.e. favourable cases will be 7×(8 rows + 8 columns) = 112.
  Hence Required Probability=(
  Favourable cases
  Sample Space
  )
  =
  112
  64C2
  
  = 1/18
• 8 years agoHelpfull: Yes(0) No(0)
• if we choose a particular square then the number of ways for choosing another square on the same side are : 7 + 7 = 14
  choosing such a pair in whole chessboard = 8 * 14 = 112
  now probability = 112/64c2 = 1/8.
• 5 years agoHelpfull: Yes(0) No(0)
• Total number of squares = 64
  
  Two squares can be selected in 64C2 ways
  
  In each column, there are 7 pairs of adjacent squares where each pair share 1 side in common.
  Total such pairs = 8×7 = 56
  
  In each row, there are 7 pairs of adjacent squares where each pair share 1 side in common.
  Total such pairs = 8×7 = 56
  
  Therefore, favourable cases = 56 + 56 = 112
  
  Required probability
  =
  112/
  64
  C
  2
  =
  112/
  2016
  =
  1/
  18
• 5 years agoHelpfull: Yes(0) No(0)