Two squares of `1xx1` are chosen at random on a chestboard. What is the probability that they have a side in common ?
A
`1//18`
B
`64//4032`
C
`63//64`
D
`1//9`
Text Solution
Verified by Experts
The correct Answer is:
A
`(a)` Total number of ways of selecting two squares `"^(64)C_(2)=(1)/(2)(64xx63)=2016` 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 required number of combinations `=(1)/(2)((4xx2)+(24xx3)+(36xx4))=112` Therefore, the required proability `=1//18`.
