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2023 AMC 10B Problems/Problem 21

Revision as of 16:47, 15 November 2023 by Kjljixx (talk | contribs) (Add another solution)

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 1

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

Solution 2

We will start with all the balls outside of the boxes, and distribute them as follows:

We put $x$ balls into the first box. There is (obviously) a roughly $\frac{1}{2}$ probability $x$ is odd (It's okay to not use the exact probability since the problem asks for the closest answer choice, and the answer choices aren't very close to each other).

We put $y$ balls into the second box. There is also a roughly $\frac{1}{2}$ probability $y$ is odd.

If both $x$ and $y$ are odd, then the number of balls which go into the third box must also be odd, since 2023 is odd. Additionally, $x$ and $y$ clearly must both be odd in order for the problem conditions to be satisfied.

Therefore our answer is the probability both $x$ and $y$ are odd, which is approximately $\frac{1}{2}\cdot\frac{1}{2}=\boxed{\textbf{(E) }\dfrac14.}$

~kjljixx

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