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?
. . . $\displaystyle (A)\;\tfrac{1}{18}$
. . . $\displaystyle (B)\;\tfrac{64}{4032}$ . Why isn't this reduced?
. . . $\displaystyle (C)\;\tfrac{63}{64}$
. . . $\displaystyle (D)\;\tfrac{1}{9}$
There are: .$\displaystyle {64\choose2}\,=\,2016$ possible outcomes.
Consider the top row of the chessboard: .$\displaystyle \square\!\square\!\square\!\square\!\square\! \square\!\square\!\square $
Consider placing a horizontal domino $\displaystyle \square\!\square$ in the row.
. . There are 7 possible positions in each row.
With 8 rows, there are $\displaystyle 7\times 8 \,=\,56$ ways to place a horizontal domino.
Now consider placing a vertical domino: $\displaystyle \begin{array}{c}\square \\[-2mm] \square\end{array}$ in a column.
. . There are 7 possible positions in each column.
with 8 columns, there are $\displaystyle 7\times8 \,=\,56$ ways to place a vertical domino.
Hence, there are $\displaystyle 56 + 56 \,=\,112$ ways to place a domino.
The probability is: .$\displaystyle \frac{112}{2016} \:=\:\frac{1}{18}$ . Answer (A)