A box contains $p$ white balls and $q$ black balls. Beside the box there is a pile of black balls. Two balls are taken out of the box. If they have the same color, a black ball from the pile is put into the box. If they have different colors, the white ball is put back into the box. This procedure is repeated until the last two balls are removed from the box and one last ball is put in. What is the probability that this last ball is white?
Problem
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Tags: probability, combinatorics, counting, IMO Shortlist, IMO Longlist
MellowMelon
10.09.2010 20:46
Perhaps I'm misunderstanding something, but the parity of the number of white balls in the box never changes, so the probability is $1$ or $0$ depending on the parity of $p$.
Amir Hossein
10.09.2010 21:05
This is IMO 1982 ShortList 1982 Problem 10. Official solution: If the two balls taken from the box are both white, then the number of white balls decreases by two; otherwise, it remains unchanged. Hence the parity of the number of white balls does not change during the procedure. Therefore if $p$ is even, the last ball cannot be white; the probability is $0$. If $p$ is odd, the last ball has to be white; the probability is $1.$
FibonacciMoose
21.02.2021 19:42
Shouldn't it then be $\frac{1}{2}$, assuming that the parity of white balls is random?
hansenhe
21.02.2021 19:50
whoa what a bump
FibonacciMoose
21.02.2021 19:56
By the way, what does "bump" mean? Thank You
Ultraman
21.02.2021 20:20
FibonacciMoose wrote: Shouldn't it then be $\frac{1}{2}$, assuming that the parity of white balls is random? It is not given that $p$ is randomized, it's just a variable
FibonacciMoose
21.02.2021 20:23
OK, thanks