# 2019 AMC 12B Problems/Problem 13

The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page.

## Problem

A red ball and a green ball are randomly and independently tossed into bins numbered with the 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?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$

## Solution

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. The probability of both landing in the same bin is $\sum_{k=1}^{\infty}2^{-2k}$. The sum is equal to $\frac{1}{3}$. Therefore the other two probabilities have to both be $\textbf{(C) } \frac{1}{3}$.
$Q.E.D\blacksquare$
Solution by a1b2

## Solution 2 (variant)

Suppose the green ball goes in bin $i$, for some $i \ge 1$. The probability of this occurring is $\frac{1}{2^i}$. Given this occurs, the probability that the red ball goes in a higher-numbered bin is $\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}$. Thus the probability that the green ball goes in bin $i$, and the red ball goes in a bin greater than $i$, is $\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}$. Summing from $i=1$ to infinity, we get

$$\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}$$

(Note: to find this sum, we use the formula $\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}$. Since in this case, $r = \frac{1}{4}$, the answer is $\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$. If you don't know this formula, you may instead note that if you multiply the sum by $4$, it is equivalent to adding $1$. Thus: $4n = n+1$, which clearly simplifies to $n = \frac{1}{3}$.

- scrabbler94 (explanation of infinite sum provided by Robin)