Fundamentals of Probability Suppose I have two boxes , containing white and black balls. In the first one , we have 8 white and 6 black balls. In the second one , we have 4 white and 7 black balls.
Now if one ball is drawn at random , suppose we need to find the probability o it being black.
Now by the classical definition : there are 7 + 6 ways in which we can select a single black ball.
And the total number of ways in which we can select a single ball is 14 + 11 = 25
So the probability using this approach is 13/25
However , if we break the problem up into two parts , the probability of selecting a particular box , and then selecting a single ball , we get
(1/2* 6/14) + (1/2*7/11)
These two approaches lead to different answers.
I'd like to know why the first one is incorrect
Thanks!
 A: It all depends on whether you give equal weighting to a box, or to a ball - the question is not well framed. What does it mean to choose a ball "at random" in this case?
Your "classical" approach is valid if choosing a ball at random means "each ball has equal probability of being picked", more or less ignoring the boxes.
Your second approach is valid if choosing a ball at random means selecting one of the boxes at probability 0.5 each, and only then choosing a ball (each with equal probabilities) from within that box.
The two interpretations of "pick a ball at random" lead to different answers.
This may be made more obvious by considering the following more extreme example: the first box contains 99 white balls, the second contains a single black ball. What is the probability of choosing a black ball? Answer: it depends on your method of choosing a ball "at random", ie whether you assign equal probabilities to boxes, or to balls.  
A: Once you say that there are two boxes, you obviously can draw a ball from only one box, so lumping together all the balls as if there were no boxes is incorrect.
Assuming that each box has equal probability of being picked, the 2nd approach is correct.
