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| − | ==Problem==
| + | #redirect[[2023 AMC 12B Problems/Problem 19]] |
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| − | 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?
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| − | == Solution 1 ==
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| − | We first examine the possible arrangements for parity of number of balls in each box for <math>2022</math> balls.
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| − | If a <math>0</math> denotes an even number and a <math>1</math> denotes an odd number, then the distribution of balls for <math>2022</math> balls could be <math>000,011,101,</math> or <math>110</math>. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
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| − | From <math>000</math>, it is not possible to get to all odd by adding one ball; we could either get <math>100,010,</math> or <math>001</math>. For the other <math>3</math> cases, though, if we add a ball to the exact right place, then it'll work.
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| − | For each of the working cases, we have <math>1</math> possible slot the ball can go into (for <math>101</math>, for example, the new ball must go in the center slot to make <math>111</math>) out of the <math>3</math> slots, so there's a <math>\dfrac13</math> chance. We have a <math>\dfrac34</math> chance of getting one of these working cases, so our answer is <math>\dfrac34\cdot\dfrac13=\boxed{\textbf{(E) }\dfrac14.}</math>
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| − | ~Technodoggo
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| − | == Solution 2 ==
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| − | We will start with all the balls outside of the boxes, and distribute them as follows:
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| − | We put <math>x</math> balls into the first box. There is (obviously) a roughly <math>\frac{1}{2}</math> probability <math>x</math> 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).
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| − | We put <math>y</math> balls into the second box. There is also a roughly <math>\frac{1}{2}</math> probability <math>y</math> is odd.
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| − | If both <math>x</math> and <math>y</math> are odd, then the number of balls which go into the third box must also be odd, since 2023 is odd.
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| − | Additionally, <math>x</math> and <math>y</math> clearly must both be odd in order for the problem conditions to be satisfied.
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| − | Therefore our answer is the probability both <math>x</math> and <math>y</math> are odd, which is approximately <math>\frac{1}{2}\cdot\frac{1}{2}=\boxed{\textbf{(E) }\dfrac14.}</math>
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| − | ~kjljixx
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| − | ==Solution 3==
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| − | We use the generating functions approach to solve this problem.
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| − | Define <math>\Delta = \left\{ \left( a, b, c \right) \in \Bbb Z_+: a+b+c = 2023 \right\}</math>.
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| − | We have
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| − | <cmath>
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| − | \[
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| − | \left( x + y + z \right)^{2023}
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| − | = \sum_{(a,b,c) \in \Delta}
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| − | \binom{2023}{a,b,c} x^a y^b z^c .
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| − | \]
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| − | </cmath>
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| − | First, we set <math>x \leftarrow 1</math>, <math>y \leftarrow 1</math>, <math>z \leftarrow 1</math>.
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| − | We get
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| − | <cmath>
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| − | \[
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| − | 3^{2023}
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| − | = \sum_{(a,b,c) \in \Delta}
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| − | \binom{2023}{a,b,c} 1 . \hspace{1cm} (1)
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| − | \]
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| − | </cmath>
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| − | Second, we set <math>x \leftarrow 1</math>, <math>y \leftarrow -1</math>, <math>z \leftarrow 1</math>.
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| − | We get
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| − | <cmath>
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| − | \[
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| − | 1
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| − | = \sum_{(a,b,c) \in \Delta}
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| − | \binom{2023}{a,b,c} (-1)^b . \hspace{1cm} (2)
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| − | \]
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| − | </cmath>
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| − | Third, we set <math>x \leftarrow 1</math>, <math>y \leftarrow 1</math>, <math>z \leftarrow -1</math>.
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| − | We get
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| − | <cmath>
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| − | \[
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| − | 1
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| − | = \sum_{(a,b,c) \in \Delta}
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| − | \binom{2023}{a,b,c} (-1)^c . \hspace{1cm} (3)
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| − | \]
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| − | </cmath>
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| − | Fourth, we set <math>x \leftarrow 1</math>, <math>y \leftarrow -1</math>, <math>z \leftarrow -1</math>.
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| − | We get
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| − | <cmath>
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| − | \[
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| − | -1
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| − | = \sum_{(a,b,c) \in \Delta}
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| − | \binom{2023}{a,b,c} (-1)^{b+c} . \hspace{1cm} (4)
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| − | \]
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| − | </cmath>
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| − | Taking <math>\frac{(1)-(2) - (3)+(4)}{4}</math>, we get
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| − | <cmath>
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| − | \begin{align*}
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| − | \frac{3^{2023} - 1 - 1 + (-1)}{4}
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| − | & = \frac{1}{4}
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| − | \sum_{(a,b,c) \in \Delta}
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| − | \binom{2023}{a,b,c}
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| − | \left(
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| − | 1 - (-1)^b - (-1)^c + (-1)^{b+c}
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| − | \right) \\
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| − | & = \frac{1}{4}
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| − | \sum_{(a,b,c) \in \Delta}
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| − | \binom{2023}{a,b,c}
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| − | \left( 1 - (-1)^b \right)
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| − | \left( 1 - (-1)^c \right) \\
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| − | & = \sum_{\substack{(a,b,c) \in \Delta \\ a, b, c \mbox{ are odds}}}
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| − | \binom{2023}{a,b,c} .
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| − | \end{align*}
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| − | </cmath>
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| − | The last expression above is the number of ways to get all three bins with odd numbers of balls.
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| − | Therefore, this happens with probability
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| − | <cmath>
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| − | \begin{align*}
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| − | \frac{\frac{3^{2023} - 1 - 1 + (-1)}{4}}{3^{2023}}
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| − | & \approx \boxed{\textbf{(E) } \frac{1}{4}}.
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| − | \end{align*}
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| − | </cmath>
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| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | ==Solution 4==
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| − | First, we find the number of ways we can split <math>2023</math> into the sum of <math>3</math> odd numbers. Because each bin must have an odd number of balls, we can set up the equation
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| − | <cmath>
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| − | \begin{align*}
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| − | (2a+1)+(2b+1)+(2c+1)=2023
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| − | \end{align*}
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| − | </cmath>
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| − | for positive integers <math>a,b,c.</math>
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| − | Simplifying, we get <math>a+b+c=1010</math>, which we can represent using stars and bars as
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| − | <cmath>\binom{1010+3-1}{3-1}=\binom{1012}{2}=\frac{1012 \cdot 1011}{2}=506 \cdot 1011.</cmath>
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| − | By stars and bars, the total number of ways to put <math>2023</math> balls into <math>3</math> bins with no restrictions is
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| − | <cmath>\binom{2023+3-1}{3-1}=\binom{2025}{2}=\frac{2025 \cdot 2024}{2}=2025 \cdot 1012.</cmath>
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| − | Therefore, the probability that all bins contain an odd number of balls is <math>\frac{506 \cdot 1011}{2025 \cdot 1012}.</math> Since <math>\frac{506}{2025}\approx \frac{1}{4}</math> and <math>\frac{1011}{1012}\approx 1,</math> this is approximately equal to <math>\frac{1}{4} \cdot 1=\boxed{\textbf{(E) } \frac{1}{4}}.</math>
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| − | ~sweatguy
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| − | ==Solution 5==
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| − | 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.
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| − | Since there are <math>{2025 \choose 2023}</math> ways to choose the balls in total, our answer is
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| − | <math>\frac{{1012 \choose 1010}}{{2025 \choose 2023}}</math>
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| − | <math>=\frac{1012!\cdot2023!}{1010!\cdot2025!}</math>
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| − | <math>=\frac{1012\cdot1011}{2024\cdot2025}</math>
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| − | <math>\approx\boxed{\textbf{(E) }\frac{1}{4}}</math>
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