Probability - Quant/Math - CAT 2008
Question 4 the day:
October 25, 2002
The question for the day is from the topic of Probability.
- Two squares are chosen at random on a chessboard. What is the
probability that they have a side in common?
(1) |
1 / 18 |
|
(2) |
64 / 4032 |
|
(3) |
63 / 64 |
|
(4) |
1 / 9 |
Correct Answer - (1)
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 .