You have two boxes, and 100 balls-- 50 of them are red and 50 are blue. You have to distribute those balls into those two boxes in such a manner that, if you select a box randomly, and then select a ball randomly from that box, the chance of that ball being red would be maximum. (There is no restriction over the distribution process, you can even put all the 100 balls in a single box leaving the other box empty !!) ?
19-Ans- Obviously, if we equally distribute all the balls in the two boxes (box A has red:blue = 25:25, box B also has red:blue = 25:25), then the chance of choosing a box randomly and finally choosing a red ball from that box randomly would be : [(1/2)(25/50) + (1/2)(25/50)] = 1/2.
Now, the question is can we increase this chance ?
What if we put all the balls in box A and leave box B empty ??
Then the final chance would be: [(1/2)(50/100) + (1/2)(0)] = 1/4.
But what if we put 99 balls in box A and only 1 red ball in box B ??
Then the chance would be : [(1/2)(49/99) + (1/2)(1/1)] = 3/4 (approx.)
If we put 98 balls in A and 2 red balls in B, then the chance would become:
[(1/2)(48/98) + (1/2)(2/2)] < [(1/2)(49/99) + (1/2)(1/1)].
Similarly it can be shown that all other solutions will yield lower chances.
Thus the best distribution could be 99 balls in one box, 1 red ball in another box.