The probability of either the Rook or the Bishop attacking the other is 0.3611
A Rook and a Bishop on a standard chess-board can be arranged in 64P2 = 64*63 = 4032 ways
Now, there are 2 cases - Rook attacking Bishop and Bishop attacking Rook. Note that the Rook and the Bishop never attack each other simultaneously. Let's consider both the cases one by one.
Case I - Rook attacking Bishop
The Rook can be placed in any of the given 64 positions and it always attacks 14 positions. Hence, total possible ways of the Rook attacking the Bishop = 64*14 = 896 ways
Case II - Bishop attacking Rook
View the chess-board as a 4 co-centric hollow squares with the outermost square with side 8 units and the innermost square with side 2 units.
If the bishop is in one of the outer 28 squares, then it can attack 7 positions. If the bishop is in one of the 20 squares at next inner-level, then it can attack 9 positions. Similarly if the bishop is in one of the 12 squares at next inner-level, then it can attack 11 positions. And if the bishop is in one of the 4 squares at next inner-level (the innermost level), then it can attack 13 positions.
Hence, total possible ways of the Bishop attacking the Rook
= 28*7 + 20*9 + 12*11 + 4*13
= 560 ways
Thus, the required probability is
= (896 + 560) / 4032
= 13/36
= 0.3611
