Difference between revisions of "2023 AMC 12B Problems/Problem 19"
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For case 1 which has a probability <math>p(n-1)</math>, in order to get all 3 bins to be odd, we need put the last 2 balls in the same bin. This has a probability of <math>\frac{1}{3}</math>. | For case 1 which has a probability <math>p(n-1)</math>, in order to get all 3 bins to be odd, we need put the last 2 balls in the same bin. This has a probability of <math>\frac{1}{3}</math>. | ||
− | For case 2 which has probability of <math>1-p(n-1)</math>, 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 <math>\frac{2}{3}</math> to be put in an even bin, and the second has a probability of <math>\frac{1}{3}</math> to be put the in the last even bin, so the probability is <math>\frac{2}{9}</math> to get all 3 bins to be odd in this case. | + | For case 2 which has a probability of <math>1-p(n-1)</math>, 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 <math>\frac{2}{3}</math> to be put in an even bin, and the second has a probability of <math>\frac{1}{3}</math> to be put the in the last even bin, so the probability is <math>\frac{2}{9}</math> to get all 3 bins to be odd in this case. |
So we have <math>p(n)=\frac{1}{3} p(n-1) + \frac{2}{9} (1-p(n-1)) = \frac{1}{9} p(n-1) + \frac{2}{9}</math>. | So we have <math>p(n)=\frac{1}{3} p(n-1) + \frac{2}{9} (1-p(n-1)) = \frac{1}{9} p(n-1) + \frac{2}{9}</math>. |
Revision as of 14:24, 20 November 2023
- 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 3
- 6 Solution 4
- 7 Solution 5
- 8 Solution 6
- 9 Solution 7
- 10 Solution 8
- 11 Solution 9 (Elementary)
- 12 Solution 10 (Exact Probability)
- 13 Video Solution 1 by OmegaLearn
- 14 Video Solution 2 by paixiao
- 15 Video Solution
- 16 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.
Solutions 8 and 9 both incorrectly state that the probability of an odd number of balls in the first bin is .
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 misleading and poorly explained 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 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
.
~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 3
Having 2 bins with an odd number of balls means the 3rd bin also has an odd number. The probability of the first bin having an odd number of balls is , since even and odd have roughly the same probability. The probability of the second bin having an odd number of balls is also
for the same reason. If both of these bins have an odd number of balls, the number of balls remaining for the third bin is also odd. Therefore the probability is
.
~Yash C
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
Really simple way to solve it. To have 3 numbers that are all odd, you need to get odd for the first two bins, and the last one will always be odd. There are 2023 ball, so the chance of having a odd number in the first bin is 1012/2023 and the chance of having another odd is 1/2. 1012/2023 * 1/2 is closest to 1/4.
~Jack Bai
Solution 9 (Elementary)
For the first bin we have options to have an odd amount of balls.
By simple elementary rules, we see the remaining amount of balls is even, and therefore we have a chance of the amount of balls in the second bin being odd.
The amount of balls in the last bin is just set by the amount of balls in the previous bins, by intuition.
Therefore we have and so it is just
.
~ESAOPS
Solution 10 (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 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
.
Video Solution 1 by OmegaLearn
Video Solution 2 by paixiao
https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=403s
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.