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 from the
box.
If they are of the same colour, a black ball from the pile is put into
the box.
If they are of different colours, the white ball is put back into the box.
This procedure is repeated until the last pair of balls are removed from the box and one last ball is put in.
What is the probability that this last ball is white?
Source: Australian Olympiad 1983
In every round of this process, two balls are removed and one is placed in, decreasing the number by one on net.
If two white balls are removed and one black one added, the count of black balls increases by one.
If two black balls are removed and one added back in, the count of black balls decreases by one.
If one of each color is removed and one white ball added, then the count of black balls decreases by one.
In each case, the parity of the number of black balls changes, from odd to even or even to odd; on no round does the parity stay the same.
For the last ball to be white, the last round must change the parity of the number of black balls to even. This happens if that parity was even to begin with and there were an even number of steps (that is, there were an odd number of balls all together) or the black-ball parity was odd to begin with and there were an odd number of steps (an even total of black and white balls).
The above happens if there were an odd number, p, of white balls.
That makes the probability that the last ball is white equal 1 if p is odd, and 0 if p is even.
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Posted by Charlie
on 2018-10-16 11:33:43 |
solution
