Answer / guest

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

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