Question on probability there are two bowls with black olives in one and green in the other. A boy takes 20 green olives and puts in the black olive bowl, mixes the black olive bowl, takes 20 olives and puts it in the green olive bowl. The question is -
Are there more green olives in the black olive bowl or black olive in the green olive bowl? Answer with reason.
 A: There are just as many green olives in the black olive bowl as there are black olives in the green olive bowl. The number of olives in each bowl hasn't changed; hence there has merely been an exchange of some black olives for an equal number of green olives.
A: After the two transfers, the number of olives in the left-hand bowl is unchanged, and the number of olives in the right-hand bowl is unchanged. 
So if we look at the left-hand bowl ("green olive bowl") any black olives that end up there must displace exactly as many green olives. Since, implausibly, none are eaten by the transferer, they must end up in the right-hand bowl.  
A: Suppose you start with $B$ black olives in one bowl and $G$ green olives in the other.
After the first transfer, we have $B$ black olives and $20$ green olives in one bowl and $G-20$ green olives in the other.
For the second transfer, suppose the boy picks $X$ black olives and $20-X$ green olives (for a total of $20$ olives).
Then:
In the black olive bowl, we have $B - X$ black olives and $20-(20-X) = X$ green olives.
In the green olive bowl, we have $X$ black olives and $(G-20)+(20-X) = G-X$ green olives.
Thus we end up with the same number of green olives in the black olive bowl as black olives in the green olive bowl (namely $X$).
(Or in table form, with $(A,B)$ denoting the number of black and green olives respectively):
Black bowl | Green bowl
(B,0)      | (0,G)    
(B,20)     | (0,G-20)      Transfer 20 green, from right to left    
(B-X,X)    | (X,G-X)       Transfer X black and 20-X green, from left to right

