2023 AMC 12B Problems/Problem 19
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
Solution 2
Since 2023 is an odd number, the 3 bins can only be
We want to find the probability that all bins are odd and it can be seen that it is approximately a chance. .
~pengf
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
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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All AMC 12 Problems and Solutions |
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