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.
Contents
- 1 Problem
- 2 Important Clarification
- 3 Solution 1
- 4 Solution 2 (Systematic Algebraic Way)
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Solution 7
- 9 Solution 8 (Exact Probability)
- 10 Solution 9 (Recursion, rigorous)
- 11 Solution 10 (Generating Functions instead of stars and bars)
- 12 Solution 11(Close and Simple)
- 13 Video Solution 0 by MegaMath
- 14 Video Solution 1 by OmegaLearn
- 15 Video Solution
- 16 Video Solution 4 by Lucas637
- 17 See also
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?
Important Clarification
Stars and Bars does not provide an exact probability. However, it does provide a good estimate for the approximate answer as on average, the number of arrangements will be almost the same when each container has an odd # of balls or when each container has an even # of balls. (similar binomial distributions)
The above clarification is a good explanation of why the stars and bars argument still yields an expression sufficiently closed to . However, I think the content of this clarification should be clearly explained WITHIN any stars and bars solution. Even with this clarification, this page is currently a redundant mess of poorly explained fakesolves:
Solutions 1 and 2 both use stars and bars to arrive at an incorrect probability and fail to explain why their method is a good approximation of the actual probability.
The explanation in solution 4 does not make sense at all.
The nature of this problem makes it so that pretty much any approach will arrive at an answer close to . However, I'm concerned that the current state of this page will make most readers come away with a worsened understanding of the structure of the problem, especially since the first two solutions are both fakesolves. This page should be fixed ASAP.
Please don't delete this message without addressing any of my concerns. If you have any questions, feel free to PM me. ~ CT17
(Solution 5 by Dissmo does a very good job at explaining Solution 4.) -Multpi12
Is the reason why the answers aren't exact because the bins are not distinct? Is it because of the situations where some bins have the same number of balls? ~ Bread10
The reason why solutions using stars and bars are not exact is because they make incorrect assumptions about the probability space. In the stars and bars solution, the sample space is the set of all orderings such that are nonnegative integers satisfying . However, not every element in this sample space is equally likely, so one cannot calculate probability by simply dividing the # of favorable outcomes over the # of total outcomes (# of elements in the sample space). ~tsun26
Solution 1
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 .
Note: My solution does not provide an exact probability, but is a good estimate, which is how this problem was designed. Most of the solutions on this page will give a decent estimate. This is due to the fact that the binomial expansions are symmetric for even vs odd. Try experimenting to determine why this is true.
~lprado
~AtharvNaphade ~eevee9406 ~Teddybear0629
Solution 2 (Systematic Algebraic Way)
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 . We can also see that , , 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 =.
Now we need to calculate /, which is just . Cancelling the twos, we get . This is roughly equal to . The answer is .
~Aopsthedude
Solution 4
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 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:
(,,)
()
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 6
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)
Solution 7
Four even-odd splittings divides in to three, namely , , , and . Here if we define a "move" as relocated one ball, then we will notice in each case, that a random "move" will be evenly likely to be one of the other three splittings. Hence by Group theory (or by intuition), we will find the structure of this splitting is 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 .
~Prof. Joker
Solution 8 (Exact Probability)
Let be the probability of all 3 bins having odd number of balls when we randomly put balls into these bins. We need find out . Clearly . For , there are only two scenarios for the first balls: Case 1 all 3 bins are odd; Case 2 only 1 bin is odd and the other 2 bins are even.
For case 1 which has a probability , in order to get all 3 bins to be odd, we need to put the last 2 balls in the same bin. This has a probability of .
For case 2 which has a probability of , in order to get all 3 bins to be odd, we need to put the last 2 balls into the 2 even bins. The first one has a probability of to be put in an even bin, and the second has a probability of to be put the in the last even bin, so the probability is to get all 3 bins to be odd in this case.
So we have .
Solving for the fixed point: yields , and this helps us the rewrite the above as .
Therefore . Using , we have .
The probability for 2023 balls is , and this is closest to .
~Qing
Solution 9 (Recursion, rigorous)
We can use recursion to figure out an expression that determines the probability of all the bins being odd.
Define to be the probability that with balls, randomly inserting them makes all 3 bins have an odd count. We would first want to find for some function .
Now, we can say that if we have an O-O-O combination, where O denotes odd and E denotes even, then to add 2 more balls to keep it O-O-O, we need to put them in the same bin. This would result in a probability of staying O-O-O.
If we have an E-E-O combination, then we need to add 2 more balls, one in each of the even bins. This would result in a probability of turning to O-O-O. Now, we can write a function of in terms of .
Now, we want to express for some function . Here, the is a problem because now we can't just put it as a geometric sequence. However, we can rewrite it as followed:
Now, we can let . We can rewrite that as .
We can note that . Since (you only have one ball, so two of the bins have to be even), . Therefore, .
Substituting, we can find that .
2023 is the 1012th odd number, so we can substitute into the equation.
.
Therefore, the answer is
~ethanzhang1001
Solution 10 (Generating Functions instead of stars and bars)
Define
Similarly we can compute the total number of cases which is:
So, the probability is:
Therefore, the answer is
Solution 11(Close and Simple)
By Stars and Bars, the total number of ways to place the 2023 balls into 3 distinct bins is 2025C2.
Let's do some case work now.
If Bin A has 1 ball, Bin 2 and Bin 3 have how 2022 balls total.
If Bin 2 has 1 ball, bin 3 has 2021.
If Bin 2 has 3 balls, Bin 3 has 2019 balls.
So, in total, there are ways for case 1.
By symmatry, if Bin A now has 3 balls, there are 1010 ways. Continuing down to if Bin A has 2021 balls, we have the total amount of successful ways is 1011+1010+1009+....+1. So the probability is closest to , which gives us .
~MC
Video Solution 0 by MegaMath
https://www.youtube.com/watch?v=aVkRJLbcD2o&t=1s ~not megaehertz I hope
Basically Solution 1
Video Solution 1 by OmegaLearn
Basically Solution 1
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 4 by Lucas637
https://www.youtube.com/watch?v=qZRQxiPIRhI
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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