# probability

• Oct 17th 2012, 09:18 AM
lemini
probability
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, 09:27 AM
relk
Re: probability

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, 10:37 AM
Soroban
Re: 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?

. . . $(A)\;\tfrac{1}{18}$

. . . $(B)\;\tfrac{64}{4032}$ . Why isn't this reduced?

. . . $(C)\;\tfrac{63}{64}$

. . . $(D)\;\tfrac{1}{9}$

There are: . ${64\choose2}\,=\,2016$ possible outcomes.

Consider the top row of the chessboard: . $\square\!\square\!\square\!\square\!\square\! \square\!\square\!\square$

Consider placing a horizontal domino $\square\!\square$ in the row.
. . There are 7 possible positions in each row.
With 8 rows, there are $7\times 8 \,=\,56$ ways to place a horizontal domino.

Now consider placing a vertical domino: $\begin{array}{c}\square \\[-2mm] \square\end{array}$ in a column.
. . There are 7 possible positions in each column.
with 8 columns, there are $7\times8 \,=\,56$ ways to place a vertical domino.

Hence, there are $56 + 56 \,=\,112$ ways to place a domino.

The probability is: . $\frac{112}{2016} \:=\:\frac{1}{18}$ . Answer (A)