Question

Nine squares are chosen at random on a chessboard. What is the probability that they form a square of size $3\times 3$? 

• A. $\displaystyle \frac{9}{\binom{64}{9}}$
• B. $\displaystyle \frac{36}{\binom{64}{9}}$
• C. $\displaystyle \frac{6}{\binom{64}{9}}$
• D. None of these

Hint:

Number of $k\times k$ squares on an $n\times n$ board is $(n-k+1)^2$.

Answer 1

The Correct Option is B

There are $\binom{64}{9}$ ways to choose any $9$ squares from an $8\times 8$ board. A $3\times 3$ block can start in $6$ positions horizontally and $6$ vertically, so there are $6\times 6=36$ such blocks. To “form a $3\times 3$ square,” the chosen $9$ squares must be exactly one of these blocks. Hence \[ P=\frac{\text{favourable}}{\text{total}}=\frac{36}{\binom{64}{9}}. \]