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

Step 1: Total ways to pick any \(9\) squares from the \(64\) squares of the board is \(\binom{64}{9}\).

Step 2: A \(3\times 3\) block fits inside the \(8\times 8\) board only if its top-left square lies in one of \(8-3+1=6\) rows and \(6\) columns, giving \(6\times 6=36\) possible blocks.

Step 3: Probability of the chosen \(9\) squares forming such a block: \[ P=\boxed{\dfrac{36}{\binom{64}{9}}} \]