A queen, a rook and a bishop are randomly placed on distinct squares of an ordinary chessboard.
Find the probability that:.
(i) The queen is under attack from either the bishop or the rook.
(ii) The bishop is neither under attack from the queen, nor under attack from the rook.
(In reply to re(2): computer aided solution
by Steve Herman)
"The probability that the Q is in the same diagonal as the bishop is 10/63.
"The probability that the Q is on the same rank or file as the rook is 14/63.
"If they could attack through each other, than the answer would be 10/63 + 14/63 = 24/63 = 8/21 ~~ 38%."
The queen could be under attack simultaneously from bishop and rook. This would be counted twice in your accounting, but should be counted only once. P(B or R) = P(B) + P(R) - P(B and R)
By modified computer program, the figures are:
In unreduced fractions:
49728/249984 + 34720/249984 - 7840/249984 = 76608/249984
37/186 + 5/36 - 35/1116 = 19/62 ~= 30.6%
where the rook figure comes first, then the bishop figure and the double attack is subtracted out.
Edited on August 30, 2016, 8:30 pm
Posted by Charlie
on 2016-08-30 19:58:28