Problem:
A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin k is 2−k for k=1,2,3,… What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
Answer Choices:
A. 41
B. 72
C. 31
D. 83
E. 73
Solution:
By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. Clearly, the probability of both landing in the same bin is ∑k=1∞2−k⋅2−k=∑k=1∞2−2k=31 (by the geometric series sum formula). Therefore, since the other two probabilities have to both the same, they have to be 21−31= (C)31.
OR
Suppose the green ball goes in bin i, for some i≥1. The probability of this occurring is 2i1. Given that this occurs, the probability that the red ball goes in a higher-numbered bin is 2i+11+2i+21+…=2i1 (by the geometric series sum formula). Thus the probability that the green ball goes in bin i, and the red ball goes in a bin greater than i, is (2i1)2=4i1. Summing from i=1 to infinity, we get
∑i=1∞4i1=1−4141=4341= 31
where we again used the geometric series sum formula. (Alternatively, if this sum equals n, then by writing out the terms and multiplying both sides by 4 , we see 4n=n+1, which gives n=31.)
OR
For red ball in bin k,Pr( Green Below Red )=∑i=1k−12−i(GBR) and Pr(Red in Bink=2−k(RB).
Pr(GBR∣RB)=∑k=1∞2−k∑i=1k−12−i=∑k=1∞2−k⋅21(1−1/21−(1/2)k−1)
∑k=1∞2−k1−2∑k=1∞(22)−k1⟹1−2/3= (C)31
The problems on this page are the property of the MAA's American Mathematics Competitions