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.
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[hide]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
(This solution is incorrect since all distributions are not equally likely) ~ AtharvNaphade
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 . This is equal to . The total amount of ways would also be found using stars and bars. That would be . Dividing our two quantities, we get . We can roughly cancel to get . The 2 in the numerator and denominator also cancels out, so we're left with .
~lprado ~eevee9406 ~Teddybear0629
Solution 2
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
~pengf ~Technodoggo
Solution 3
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:
(,,)
()
Let the probability of achieving the case to be and any of the permutations to be .
Because the amount of balls is arbitrarily large, even after another two balls are be placed.
There are two cases for which placing another two balls results in :
: The two balls are placed in the same bin ()
: The two balls are placed in the two even bins ()
So,
-Dissmo
Solution 4
We use the generating functions approach to solve this problem. Define .
We have
First, we set , , . We get
Second, we set , , . We get
Third, we set , , . We get
Fourth, we set , , . We get
Taking , we get
The last expression above is the number of ways to get all three bins with odd numbers of balls. Therefore, this happens with probability
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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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