|
|
| (6 intermediate revisions by 3 users not shown) |
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect[[2023 AMC 12B Problems/Problem 19]] |
| − | | |
| − | 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 <math>2022</math> balls.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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>
| |
| − | | |
| − | ~Technodoggo
| |
| − | | |
| − | == Solution 2 ==
| |
| − | | |
| − | We will start with all the balls outside of the boxes, and distribute them as follows:
| |
| − | | |
| − | 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).
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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.
| |
| − | Additionally, <math>x</math> and <math>y</math> clearly must both be odd in order for the problem conditions to be satisfied.
| |
| − | | |
| − | 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>
| |
| − | | |
| − | ~kjljixx
| |
| − | | |
| − | ==Solution 3==
| |
| − | | |
| − | We use the generating functions approach to solve this problem.
| |
| − | Define <math>\Delta = \left\{ \left( a, b, c \right) \in \Bbb Z_+: a+b+c = 2023 \right\}</math>.
| |
| − | | |
| − | 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 <math>x \leftarrow 1</math>, <math>y \leftarrow 1</math>, <math>z \leftarrow 1</math>.
| |
| − | We get
| |
| − | <cmath>
| |
| − | \[
| |
| − | 3^{2023}
| |
| − | = \sum_{(a,b,c) \in \Delta}
| |
| − | \binom{2023}{a,b,c} 1 . \hspace{1cm} (1)
| |
| − | \]
| |
| − | </cmath>
| |
| − | | |
| − | Second, we set <math>x \leftarrow 1</math>, <math>y \leftarrow -1</math>, <math>z \leftarrow 1</math>.
| |
| − | We get
| |
| − | <cmath>
| |
| − | \[
| |
| − | 1
| |
| − | = \sum_{(a,b,c) \in \Delta}
| |
| − | \binom{2023}{a,b,c} (-1)^b . \hspace{1cm} (2)
| |
| − | \]
| |
| − | </cmath>
| |
| − | | |
| − | Third, we set <math>x \leftarrow 1</math>, <math>y \leftarrow 1</math>, <math>z \leftarrow -1</math>.
| |
| − | We get
| |
| − | <cmath>
| |
| − | \[
| |
| − | 1
| |
| − | = \sum_{(a,b,c) \in \Delta}
| |
| − | \binom{2023}{a,b,c} (-1)^c . \hspace{1cm} (3)
| |
| − | \]
| |
| − | </cmath>
| |
| − | | |
| − | Fourth, we set <math>x \leftarrow 1</math>, <math>y \leftarrow -1</math>, <math>z \leftarrow -1</math>.
| |
| − | We get
| |
| − | <cmath>
| |
| − | \[
| |
| − | -1
| |
| − | = \sum_{(a,b,c) \in \Delta}
| |
| − | \binom{2023}{a,b,c} (-1)^{b+c} . \hspace{1cm} (4)
| |
| − | \]
| |
| − | </cmath>
| |
| − | | |
| − | Taking <math>\frac{(1)-(2) - (3)+(4)}{4}</math>, we get
| |
| − | <cmath>
| |
| − | \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*}
| |
| − | </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
| |
| − | <math></math>
| |
| − | \begin{align*}
| |
| − | \frac{\frac{3^{2023} - 1 - 1 + (-1)}{4}}{3^{2023}}
| |
| − | & \approx \boxed{\textbf{(E) <math>\frac{1}{4}</math>}}.
| |
| − | \end{align*}
| |
| − | <math></math>
| |
| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| |
