2021 AMC 12B Problems/Problem 23
Contents
Problem
Three balls are randomly and independantly tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin is for More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is where and are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins and ) What is
Solution
"Evenly spaced" just means the bins form an arithmetic sequence.
Suppose the middle bin in the sequence is . There are different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these bins are chosen is , so the probability is the middle bin is . Then, we want the sum The answer is
Solution 2
As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be and the common difference be . Further note that each pair uniquely determines a set of 3 bins.
We have because the leftmost bin in the sequence can be any bin, and , because the bins must be distinct.
This gives us the following sum for the probability: Therefore the answer is , which is choice (A).
-Darren Yao
Solution 3
This is a slightly messier variant of solution 2. If the first ball is in bin and the second ball is in bin , then the third ball is in bin . Thus the probability is
Solution 4 (Table: Detailed Explanation of Solution 2)
Since three balls have permutations, the requested probability is by infinite geometric series, from which the answer is
~MRENTHUSIASM
Video Solution Using infinite Geometric Series
https://youtu.be/3B-3_nOTIu4 ~hippopotamus1
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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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