Difference between revisions of "2023 AMC 12B Problems/Problem 19"

Revision as of 22:08, 16 November 2023

The following problem is from both the 2023 AMC 10B #21 and 2023 AMC 12B #19, so both problems redirect to this page.

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?

$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \frac{1}{4}$

Solution 1

Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to $\binom{1012}{2}$ . This is equal to $\frac{1012\cdot1011}{2}$ . The total amount of ways would also be found using stars and bars. That would be $\binom{2023+3-1}{3-1} = \binom{2025}{2}$ . Dividing our two quantities, we get $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$ . We can roughly cancel $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ to get $\frac{1}{4}$ . The 2 in the numerator and denominator also cancels out, so we're left with $\boxed{\frac{1}{4}}$ .

~lprado

~AtharvNaphade ~eevee9406 ~Teddybear0629

The above solution is completely incorrect and should probably be removed (it assumes all distributions of marble counts are equally likely) ~ CT17

Solution 2

Having 2 bins with an odd number of balls means the 3rd bin also has an odd number. The probability of the first bin having an odd number of balls is $\frac{1}{2}$ , since even and odd have roughly the same probability. The probability of the second bin having an odd number of balls is also $\frac{1}{2}$ for the same reason. If both of these bins have an odd number of balls, the number of balls remaining for the third bin is also odd. Therefore the probability is $\frac{1}{2} \cdot \frac{1}{2} = \boxed{\textbf{(E)} \frac{1}{4}}$ .

~Yash C

Solution 3

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

If a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that each case is about equally likely.

From $EEE$ , it is not possible to get to all odd by adding one ball; we could either get $OEE,EOE,$ or $EEO$ . 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 $OEO$ , for example, the new ball must go in the center slot to make $OOO$ ) 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 $\dfrac34\cdot\dfrac13=\boxed{\textbf{(E) }\dfrac14.}$

~pengf ~Technodoggo

Solution 4

2023 is an arbitrary large number. So, we proceed assuming that an arbitrarily large number of balls have been placed.

For an odd-numbered amount of balls case, the 3 bins can only be one of these 2 combinations:

$OEE$ ( $OEE$ , $EOE$ , $EEO$ )

$OOO$ ( $OOO$ )

Let the probability of achieving the $OOO$ case to be $P(OOO) = p$ and any of the $OEE$ permutations to be $P(OEE) = 1-p$ .

Because the amount of balls is arbitrarily large, $P(OOO) = p$ even after another two balls are be placed.

There are two cases for which placing another two balls results in $OOO$ :

$OOO$ : The two balls are placed in the same bin ( $OOO\to OOE\to OOO$ )

$OEE$ : The two balls are placed in the two even bins ( $OEE\to OOE \to OOO$ )

So,

$P(OOO) = P(OOO) * \frac{1}{3} + P(OEE) * \frac{2}{3} * \frac{1}{3}$

$p = p * \frac{1}{3} + (1-p) * \frac{2}{3} * \frac{1}{3}$

$\frac{8}{9}p = \frac{2}{9}$

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

~Dissmo

Solution 5

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 6

Four even-odd splittings divides $2023$ in to three, namely $(O,O,O)$ , $(E,E,O)$ , $(E,O,E)$ , and $(O,E,E)$ . Here if we define a "move" as relocated one ball, then we will notice in each case, that a random "move" will be evenly likely to be one of the other three splittings. Hence by Group theory (or by intuition), we will find the structure of this splitting is $V_4$ group, and it's symmetric for all four elements in this Group.

Thus, no matter what is the initial starting point, four cases will be evenly likely to appear when repeated many times. The answer is $\boxed{\textbf{(E)} \frac{1}{4}}$ .

~Prof. Joker

Solution 7

Really simple way to solve it. To have 3 numbers that are all odd, you need to get odd for the first two bins, and the last one will always be odd. There are 2023 ball, so the chance of having a odd number in the first bin is 1012/2023 and the chance of having another odd is 1/2. 1012/2023 * 1/2 is closest to 1/4.

~Jack Bai(only 9 years old)

This solution is also completely wrong (it assumes all marble counts in the first bin are equally likely) ~ CT17

Video Solution 1 by OmegaLearn

https://youtu.be/MCk8S8l-2EY

@@ Line 17: / Line 17: @@
 ~Teddybear0629
-The above solution is completely incorrect and should probably be removed ~ CT17
+The above solution is completely incorrect and should probably be removed (it assumes all distributions of marble counts are equally likely) ~ CT17
 == Solution 2 ==

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 10 Problems and Solutions

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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