Two squares are chosen at random from the small squares on chess board. What is the chance that the two squares have exactly one corner in common

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$\frac{1}{144}$

$\frac{3}{144}$

$\frac{7}{144}$

$\frac{5}{144}$

answer is D.

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Detailed Solution

The number of ways of selecting two small squares one after the other form the chess board is $(64) (63)$

Let $E$ be the event of getting two squares for which one corner is common

Case (i) : Let the first sqauare is corner, there are four chances, for each of these square there exists one more square having corner as common

hence, the number of favorable cases to the event $E$ is $4_{C_{1}} \cdot 1 = 4$

Case (ii) Let the first sqauare is one of the squares in the sides other than corner, there are 24 chances, for each of these square there exists two more square having corner as common

hence, the number of favorable cases to the event $E$ is $24_{C_{1}} \cdot 2 = 48$

Case (iii) Let the first sqauare is one of the squares in the middle, there are 32 chances, for each of these square there exists four more square having corner as common

hence, the number of favorable cases to the event $E$ is $32_{C_{1}} \cdot 4 = 144$

Therefore, the probability of getting the event $E$ is $\frac{196}{64 \cdot 63} = \frac{7}{16 \cdot 9} = \frac{7}{144}$