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:
Answer (A): First compute the probability that the bin numbers in which the balls land, in the order they are tossed, form an increasing arithmetic sequence; the final answer will be 6 times this number to account for all possible orders. The probability that the bin numbers satisfy this requirement with the first ball getting tossed into bin 1 is
211(221⋅231+231⋅251+241⋅271+⋯)=21n=1∑∞2n+11⋅22n+11=21n=1∑∞23n+21=21⋅251⋅1−811=561
Similarly, if the first ball is tossed into bin i, then the probability that the bin numbers satisfy this requirement is
2i1n=1∑∞23n+2i1=71⋅8i1.
Summing over all values of i gives
i=1∑∞71⋅8i1=71⋅1−8181=491
Multiplying by 6 gives the required probability as 496, so the requested sum is 6+49=55.
OR
Let Ei,d denote the event that the first ball falls in bin i, the second ball falls in bin i+d, and the third ball falls in bin i+2d, and note that P(Ei,d)=2−3i−3d. Let E be the union of the disjoint events Ei,d, then
P(E)=i=1∑∞d=1∑∞P(Ei,d)=(i=1∑∞2−3i)(d=1∑∞2−3d)=(1−8181)2=491.
The balls may fall in bins i,i+d,i+2d in any of 3!=6 orders, so the desired probability is 496 and the requested sum is 6+49=55.
The problems on this page are the property of the MAA's American Mathematics Competitions