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Question

Two squares are chosen at random on a chessboard and let p denote the probability that they have exactly one side in common. Find 36p

A
2
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B
4
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C
5
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D
1
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Solution

The correct option is A 2
In total, there are 64 squares on a chess board.

If we differentiate them with respect to the number of squares they are adjacent to, there are 3 types of squares:

Squares adjacent to 4 other squares

Out of the 64 squares, all the squares expect for the corner ones and edge ones are adjacent to 4 other squares.

There are a total of 36 of these squares.

P(selecting such a square) =3664

Then, we need the probability of selecting a square that is adjacent to this square. Let's call this event A

P(A)=463

Squares adjacent to 3 other squares

The squares that lie on the edges of the chess board (neglecting the corner ones) are adjacent to 3 other squares.

There are 24 such squares.

P(selecting such a square) = 2464
Similarly, we need the probability of selecting a square that is adjacent to this square. Let this be event B

P(B)=363

Squares adjacent to 2 other squares

The corner squares are the ones.

There are 4 such squares.

P(selecting such a square) =464

Here also, let the event of selecting an adjacent square be C,

P(C)=263

Now, solving all the three cases and adding them will give us the answer.

Answer: P(selecting a square adjacent to 4 squares)*P(A) + P(selecting a square adjacent to 3 squares)*P(B) + P(selecting a square adjacent to 2 squares)*P(C)=(3664)(463)+(2464)(363)+(464)(263)=118
therefore 36p=11836=2

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