Difference between revisions of "2023 AMC 10B Problems/Problem 21"
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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? | 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 == | + | == Solution 1 == |
We first examine the possible arrangements for parity of number of balls in each box for <math>2022</math> balls. | We first examine the possible arrangements for parity of number of balls in each box for <math>2022</math> balls. | ||
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~Technodoggo | ~Technodoggo | ||
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| + | == 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 | ||
Revision as of 16:47, 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 
balls.
If a 
denotes an even number and a 
denotes an odd number, then the distribution of balls for balls could be 
or 
. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From 
, it is not possible to get to all odd by adding one ball; we could either get 
or 
. For the other 
cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have possible slot the ball can go into (for 
, for example, the new ball must go in the center slot to make 
) out of the slots, so there's a 
chance. We have a 
chance of getting one of these working cases, so our answer is 
~Technodoggo
Solution 2
We will start with all the balls outside of the boxes, and distribute them as follows:
We put 
balls into the first box. There is (obviously) a roughly 
probability 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 
balls into the second box. There is also a roughly probability is odd.
If both and are odd, then the number of balls which go into the third box must also be odd, since 2023 is odd.
Additionally, and clearly must both be odd in order for the problem conditions to be satisfied.
Therefore our answer is the probability both and are odd, which is approximately 
~kjljixx
