2023 AMC 10B Problems/Problem 21
Solution
We first examine the possible arrangements for parity of number of balls in each box for balls.
If a denotes an even number and a
denotes an odd number, then the distribution of balls for
balls could be
or
. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From , it is not possible to get to all odd by adding one ball; we could either get
or
. For the other
cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have possible slot the ball can go into (for
, for example, the new ball must go in the center slot to make
) out of the
slots, so there's a
chance. We have a
chance of getting one of these working cases, so our answer is