There are 4 corner squares, out of a total of 64, therefore the probability of the first sqyuare being a corner is 1/16. Then there are 8 squares adjacent to a corner square so the probability of choosing one of thes3es as the second square is 8/63. So the probability of choosing two adjacent squares the first of which is a corner is 1/126.
Now repeat for edges and interior first choices.
CB
Hello, umangarora!
There's a chessboard ... regular one with 64 squares.
If two squares are selected at random,
find the probability that they have a common side.
Choosing 2 squares from the available 64 squares,
. . there are: . choices.
How many of these are "dominos"? .
In a row, , there are 7 ways to place a domino.
. . With 8 rows, there are 56 possible "horizontal" dominos.
The same is true for the columns: .56 possible "vertical" dominos.
Hence, there are: . possible dominos.
The probability is: .
Alternatively,
If a chosen square is in a corner, then the probabilty that the 2nd chosen square
is not sharing a side with it is
hence the probability that the squares have one of them in the corner,
with another not sharing a side is
There are 6(4)=24 remaining squares along the edge that are not in a corner.
The probability of another square not sharing a side with one of these is
hence the probability that the squares include one of the remaining edge squares,
but the other does not share a side with it is
There are 36 squares not on an edge.
Therefore the probability that the squares include one of these, and the 2nd one is not sharing a side with it is
Therefore, the probability that 2 squares are selected and do share a side is