Problem:
Three balls are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability it is tossed into bin i is 2−i for i=1,2,3,…. More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is qp, where p and q are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins 3,17 , and 10.) What is p+q ?
Answer Choices:
A. 55
B. 56
C. 57
D. 58
E. 59
Solution:
"Evenly spaced" just means the bins form an arithmetic sequence.
Suppose the middle bin in the sequence is x. There are x−1 different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these 3 bins are chosen is 6⋅2−3x=6⋅8x1, so the probability x is the middle bin is 6⋅8xx−1. Then, we want the sum
6∑x=2∞8xx−1=86[81+822+833⋯]
=43[(81+821+831+⋯)+(821+831+841+⋯)+⋯]
=43[71⋅(1+81+821+831+⋯)]
=43⋅498
=496
The answer is 6+49=(A)55 .
OR
As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be a and the common difference be d. Further note that each (a,d) pair uniquely determines a set of 3 bins.
We have a≥1 because the leftmost bin in the sequence can be any bin, and d≥1, because the bins must be distinct.
This gives us the following sum for the probability:
6∑a=1∞∑d=1∞2−3a−3d=6∑a=1∞∑d=1∞2−3a⋅2−3d
=6(∑a=1∞2−3a)(∑d=1∞2−3d)
=6(∑a=1∞8−a)(∑d=1∞8−d)
=6(71)(71)
=496
Therefore the answer is 6+49=(A)55 .
The problems on this page are the property of the MAA's American Mathematics Competitions