2023 AMC 10B Problems/Problem 21

Problem

Each of 2023 balls is randomly placed into one of 3 bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?

Solution

We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.

If a $0$ denotes an even number and a $1$ denotes an odd number, then the distribution of balls for $2022$ balls could be $000,011,101,$ or $110$ . With the insanely overpowered magic of cheese, we assume that each case is about equally likely.

From $000$ , it is not possible to get to all odd by adding one ball; we could either get $100,010,$ or $001$ . For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.

For each of the working cases, we have $1$ possible slot the ball can go into (for $101$ , for example, the new ball must go in the center slot to make $111$ ) out of the $3$ slots, so there's a $\dfrac13$ chance. We have a $\dfrac34$ chance of getting one of these working cases, so our answer is $\dfrac34\cdot\dfrac13=\boxed{\textbf{(E) }\dfrac14.}$

~Technodoggo

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2023 AMC 10B Problems/Problem 21

Problem

Solution