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 =  .
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