2021 AMC 12B Problems/Problem 23
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
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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