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.
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 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)
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.