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

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- Oct 17th 2012, 08:18 AMleminiprobability
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 - Oct 17th 2012, 08:27 AMrelkRe: probability
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

for practicing more questions go to Probability | Aptitude Test Solved Problems | Interview Question Answers

All d best!! - Oct 17th 2012, 09:37 AMSorobanRe: probability
Hello, lemini!

Another approach . . .

Quote:

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$\displaystyle \square\!\square$ in the row.*domino*

. . 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)