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. $\frac{9}{{}^{64}{C}_9}$
• B. $\frac{36}{{}^{64}{C}_9}$
• C. $\frac{6}{{}^{64}{C}_9}$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{36}{{}^{64}{C}_9}$

We can choose 9 squares out of 64 squares in ${}^{64}{C}_9$ ways.
Hence, exhaustive number of cases = ${}^{64}{C}_9$
From the figure, it is clear that the given square of size 3 $\times$ 3

can be formed by using four consecutive horizontal and 4 consecutive vertical lines, which can be done in ${}^6{C}_1{\times }^6{C}_1=36ways$
Basically you can make 6 square of size 3 $\times$ 3 in vertical direction and 6 squares of the size 3 $\times$ 6 in horizontal direction. Hence total 6 $\times$ 6 = 36 squares can be chosen.
$\therefore$ The required probability = $\frac{36}{{}^{64}{C}_9}$.