Q.2.17 - A First Course in Probability

Question

If 8 rooks (castles) are randomly placed on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.  

Step-by-Step Solution

Answer

The probability that no row or file contains more than one rook $9.109\times {10}^{-6}$

Step 1: Count the total number of ways to place 8 rooks

A standard chessboard has \(8 \times 8 = 64\) squares. The total number of ways to randomly place 8 indistinguishable rooks on 64 squares is:
\(\binom{64}{8} = \frac{64!}{8! \cdot 56!} = 4{,}426{,}165{,}368\)

Step 2: Count the favorable outcomes

For no two rooks to attack each other, each row and each column must contain exactly one rook. Since there are 8 rows and 8 rooks, there is exactly one rook per row.

The rook in row 1 can go in any of 8 columns, the rook in row 2 in any of the remaining 7 columns, and so on.

Number of favorable arrangements = \(8! = 40{,}320\)

Step 3: Compute the probability

\(P = \frac{8!}{\binom{64}{8}} = \frac{40{,}320}{4{,}426{,}165{,}368} \approx 9.109 \times 10^{-6}\)

The probability that none of the 8 rooks can capture any of the others is approximately \(\boxed{9.109 \times 10^{-6}}\).