Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
A. 1/18
B. 64/4032
C. 63/64
D. 1/9
Answer – (A)
Solution:
The number of ways of choosing the first square is 64. The number of ways of choosing the second square is 63.
There are a total of 64*63=4032 ways of choosing two squares.
If the first square happens to be any of the four corner ones, the second square can be chosen in 2 ways.
If the first square happens to be any of the 24 squares on the side of the chess board, the second square can be chosen in 3 ways.
If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways.
Hence the desired number of combinations:
=(4*2)+(24*3)+(36*4)=224
Therefore, the required probability =224
4032
= 1/18
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Hello, lemini!
Another approach . . .
Two squares are chosen at random on a chessboard.
What is the probability that they have a side in common?
. . .
. . . . Why isn't this reduced?
. . .
. . .
There are: . possible outcomes.
Consider the top row of the chessboard: .
Consider placing a horizontal domino in the row.
. . There are 7 possible positions in each row.
With 8 rows, there are ways to place a horizontal domino.
Now consider placing a vertical domino: in a column.
. . There are 7 possible positions in each column.
with 8 columns, there are ways to place a vertical domino.
Hence, there are ways to place a domino.
The probability is: . . Answer (A)