Probability on selecting balls 
If I have B black balls and W white balls in a bag, what is the probability that the last one I select is white?

How shall I solve this problem?
I am not sure how to make a start, is it correct to first assume $B \lt W$?
 A: The insight that you can carry is that colored balls (or, say, cards) have no preference for position.
With this insight, you should be able to get the answer in a jiffy, viz.
Whether the question is P(1st ball is white) or P(last ball is white) or anything in the middle, 
$$Pr =\frac{W}{W+B}$$
A: The number of ways to select the balls is $(B+W)!$
The number of ways to select the balls with the last one being white is $(B+W-1)!\cdot{W}$
Hence the probability that the last ball selected is white is $\dfrac{(B+W-1)!\cdot{W}}{(B+W)!}=\dfrac{W}{B+W}$
A: If you stick a number to each ball from 1 to W+B, white balls sorted first, then you have (W+B)! equally likely possible sequence of outcomes. Now, if you write all the permutations in a column, you only need to find ones with a white ball at the end. The number of permutations ,with ball 1 at the end, is (W+B-1)!. The same is applied to other white balls and we get W(W+B-1)! permutations with a white ball at the end. Using the fact that all permutations are equally likely, we divide the number of desired outcomes (a white ball at the end) by the total number of permutations to get W/(W+B) which is as if we wanted to solve the problem: "find the probability of the first ball to be white".
A: Consider the permutation of the ball, by symmetry the probability of choosing white ball as the last ball is the same as choosing the white ball as the first ball.
Thus P=$W\over {W+B}$
