Difference between revisions of "2023 AMC 10B Problems/Problem 21"

Revision as of 18:59, 15 November 2023

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

Solution 3

We use the generating functions approach to solve this problem. Define $\Delta = \left\{ \left( a, b, c \right) \in \Bbb Z_+: a+b+c = 2023 \right\}$ .

We have $\left( x + y + z \right)^{2023} = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} x^a y^b z^c .$

First, we set $x \leftarrow 1$ , $y \leftarrow 1$ , $z \leftarrow 1$ . We get $3^{2023} = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} 1 . \hspace{1cm} (1)$

Second, we set $x \leftarrow 1$ , $y \leftarrow -1$ , $z \leftarrow 1$ . We get $1 = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} (-1)^b . \hspace{1cm} (2)$

Third, we set $x \leftarrow 1$ , $y \leftarrow 1$ , $z \leftarrow -1$ . We get $1 = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} (-1)^c . \hspace{1cm} (3)$

Fourth, we set $x \leftarrow 1$ , $y \leftarrow -1$ , $z \leftarrow -1$ . We get $-1 = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} (-1)^{b+c} . \hspace{1cm} (4)$

Taking $\frac{(1)-(2) - (3)+(4)}{4}$ , we get $\begin{align*} \frac{3^{2023} - 1 - 1 + (-1)}{4} & = \frac{1}{4} \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} \left( 1 - (-1)^b - (-1)^c + (-1)^{b+c} \right) \\ & = \frac{1}{4} \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} \left( 1 - (-1)^b \right) \left( 1 - (-1)^c \right) \\ & = \sum_{\substack{(a,b,c) \in \Delta \\ a, b, c \mbox{ are odds}}} \binom{2023}{a,b,c} . \end{align*}$

The last expression above is the number of ways to get all three bins with odd numbers of balls. Therefore, this happens with probability $\begin{align*} \frac{\frac{3^{2023} - 1 - 1 + (-1)}{4}}{3^{2023}} & \approx \boxed{\textbf{(E) } \frac{1}{4}}. \end{align*}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 4

We can distribute the three odd balls to each of the bins first and divide by $2$ , leaving $1010$ balls to be distributed. This forces the number of balls in each bin to be odd. There are ${1012 \choose 1010}$ ways to sort the balls by Stars and Bars.

Since there are ${2025 \choose 2023}$ ways to choose the balls in total, our answer is

$\frac{{1012 \choose 1010}}{{2025 \choose 2023}}$

$=\frac{1012!\cdot2023!}{1010!\cdot2025!}$

$=\frac{1012\cdot1011}{2024\cdot2025}$

$\approx\boxed{\textbf{(E) }\frac{1}{4}}$

@@ Line 113: / Line 113: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Solution 4==
+We can distribute the three odd balls to each of the bins first and divide by <math>2</math>, leaving <math>1010</math> balls to be distributed. This forces the number of balls in each bin to be odd. There are <math>{1012 \choose 1010}</math> ways to sort the balls by Stars and Bars.
+Since there are <math>{2025 \choose 2023}</math> ways to choose the balls in total, our answer is
+<math>\frac{{1012 \choose 1010}}{{2025 \choose 2023}}</math>
+<math>=\frac{1012!\cdot2023!}{1010!\cdot2025!}</math>
+<math>=\frac{1012\cdot1011}{2024\cdot2025}</math>
+<math>\approx\boxed{\textbf{(E) }\frac{1}{4}}</math>

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