Conditional probability Two queens are put on a chess table. What is the probability that the queens are attacking eachother. (Conditional probability answer is asked) My thoughts; The two queens having the coordinates (i,j) and (m,n) will attack eachother if either m=i, or n=j, now i dont how to define the relation, when the diagonals are taken into consideration?
 A: Queens can attack along rows or columns. That is when $j=n \vee i=m$
Queens can attack along the diagonals too. This is when $(m−i=n−j)\vee (m−i=n+j)$ 

However to measure the probability, consider that of the $63$ spaces other than where $(i,j)$, where the white queen is at, there are $14$ squares in the same row or column. 
Now count how many other squares are in the $/$ diagonal, and count how many other squares are in the $\backslash$ diagonal.
A: The number of squares attacked by a queen along a row or column is $14$.
If you consider the chessboard carefully, I think you will see that the number of squares attacked on the diagonal depends only on the position of the queen relative to the edge of the board:


*

*from any edge square, the queen attacks $7$ squares diagonally;

*from any square one step in from the edge, she attacks $9$ squares diagonally;

*two steps in, $11$ squares;

*from one of the four centre squares, $13$ squares.


So the total number of pairs of attacking queens is
$$(28\times21)+(20\times23)+(12\times25)+(4\times27)=1456\ .\tag{$*$}$$
This assumes that the queens are placed in order, so to get the probability you divide by $P(64,2)$.  The answer is
$$\frac{1456}{64\times63}=\frac{13}{36}\ .$$
To answer the problem using conditional probability explicitly, you would want to rewrite $(*)$ as
$$\Bigl(\frac{28}{64}\times\frac{21}{63}\Bigr)+etc\ .$$
