2023 AMC 12B Problems/Problem 19
- 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 balls is randomly placed into one of
bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?
Solution 1
Suppose the numbers are ,
, and
. First, we try to calculate the amount of ways for all three balls to be placed in a bin so the number of balls in each bin is odd.
and each bin has at least one ball because they are positive odd numbers. Changing the equation, we see that
. Let
,
, and
. Thus
b_1
b_24, and
are all positive. Using the positive version of stars and bars, we get
=
choices.
Now, we want to find the total amount of cases. Using the non-negative version of stars and bars, we find that the total is =
.
/
is roughly equal to 1/4. The answer is \boxed{\textbf{(E)} \frac{1}{4}}$.
==Solution 2==
Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to$ (Error compiling LaTeX. Unknown error_msg)\binom{1012}{2}\frac{1012\cdot1011}{2}
\binom{2023+3-1}{3-1} = \binom{2025}{2}
\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}
\frac{1012 \cdot 1011}{2025 \cdot 2024}
\frac{1}{4}
\boxed{\frac{1}{4}}$.
~lprado
~AtharvNaphade ~eevee9406 ~Teddybear0629
The above solution is completely incorrect and should probably be removed (it assumes all distributions of marble counts are equally likely) ~ CT17
== Solution 3 ==
Having 2 bins with an odd number of balls means the 3rd bin also has an odd number. The probability of the first bin having an odd number of balls is$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{2}\frac{1}{2}
\frac{1}{2} \cdot \frac{1}{2} = \boxed{\textbf{(E)} \frac{1}{4}}$.
~Yash C
==Solution 4==
We first examine the possible arrangements for parity of number of balls in each box for$ (Error compiling LaTeX. Unknown error_msg)2022$balls.
If a$ (Error compiling LaTeX. Unknown error_msg)EO
2022
EEE,EOO,OEO,
OOE$. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From$ (Error compiling LaTeX. Unknown error_msg)EEEOEE,EOE,
EEO
3$cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have$ (Error compiling LaTeX. Unknown error_msg)1OEO
OOO
3
\dfrac13
\dfrac34
\dfrac34\cdot\dfrac13=\boxed{\textbf{(E) }\dfrac14.}$~pengf
~Technodoggo
==Solution 5==
2023 is an arbitrary large number. So, we proceed assuming that an arbitrarily large number of balls have been placed.
For an odd-numbered amount of balls case, the 3 bins can only be one of these 2 combinations:$ (Error compiling LaTeX. Unknown error_msg)OEEOEE
EOE
EEO
OOO
OOO$)
Let the probability of achieving the$ (Error compiling LaTeX. Unknown error_msg)OOOP(OOO) = p
OEE
P(OEE) = 1-p$.
Because the amount of balls is arbitrarily large,$ (Error compiling LaTeX. Unknown error_msg)P(OOO) = p$even after another two balls are be placed.
There are two cases for which placing another two balls results in$ (Error compiling LaTeX. Unknown error_msg)OOOOOO
OOO\to OOE\to OOO
OEE
OEE\to OOE \to OOO$)
So,$ (Error compiling LaTeX. Unknown error_msg)P(OOO) = P(OOO) * \frac{1}{3} + P(OEE) * \frac{2}{3} * \frac{1}{3}$$ (Error compiling LaTeX. Unknown error_msg)p = p * \frac{1}{3} + (1-p) * \frac{2}{3} * \frac{1}{3}$$ (Error compiling LaTeX. Unknown error_msg)\frac{8}{9}p = \frac{2}{9}$$ (Error compiling LaTeX. Unknown error_msg)p = \boxed {\textbf {(E)} \frac {1}{4}}$~Dissmo
==Solution 6==
We use the generating functions approach to solve this problem. Define$ (Error compiling LaTeX. Unknown error_msg)\Delta = \left\{ \left( a, b, c \right) \in \Bbb Z_+: a+b+c = 2023 \right\}$.
We have <cmath> \[ \left( x + y + z \right)^{2023} = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} x^a y^b z^c . \] </cmath>
First, we set$ (Error compiling LaTeX. Unknown error_msg)x \leftarrow 1y \leftarrow 1
z \leftarrow 1$.
We get
<cmath>
\[
3^{2023}
= \sum_{(a,b,c) \in \Delta}
\binom{2023}{a,b,c} 1 . \hspace{1cm} (1)
\]
</cmath>
Second, we set$ (Error compiling LaTeX. Unknown error_msg)x \leftarrow 1y \leftarrow -1
z \leftarrow 1$.
We get
<cmath>
\[
1
= \sum_{(a,b,c) \in \Delta}
\binom{2023}{a,b,c} (-1)^b . \hspace{1cm} (2)
\]
</cmath>
Third, we set$ (Error compiling LaTeX. Unknown error_msg)x \leftarrow 1y \leftarrow 1
z \leftarrow -1$.
We get
<cmath>
\[
1
= \sum_{(a,b,c) \in \Delta}
\binom{2023}{a,b,c} (-1)^c . \hspace{1cm} (3)
\]
</cmath>
Fourth, we set$ (Error compiling LaTeX. Unknown error_msg)x \leftarrow 1y \leftarrow -1
z \leftarrow -1$.
We get
<cmath>
\[
-1
= \sum_{(a,b,c) \in \Delta}
\binom{2023}{a,b,c} (-1)^{b+c} . \hspace{1cm} (4)
\]
</cmath>
Taking$ (Error compiling LaTeX. Unknown error_msg)\frac{(1)-(2) - (3)+(4)}{4}$, we get
<cmath>
The last expression above is the number of ways to get all three bins with odd numbers of balls.
Therefore, this happens with probability
<cmath>
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
==Solution 7==
Four even-odd splittings divides$ (Error compiling LaTeX. Unknown error_msg)2023(O,O,O)
(E,E,O)
(E,O,E)
(O,E,E)
V_4$group, and it's symmetric for all four elements in this Group.
Thus, no matter what is the initial starting point, four cases will be evenly likely to appear when repeated many times. The answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(E)} \frac{1}{4}}$.
~Prof. Joker
Solution 8
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
See Also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 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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