2019 AMC 12B Problems/Problem 13

Revision as of 15:45, 14 February 2019 by Expilncalc (talk | contribs) (I have transferred a solution that was put up on Problem 16 (accidentally) and put it here as "Solution Variant".)
The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page.


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?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$


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 $\textbf{(C) } \frac{1}{3}$.
Solution by a1b2

Solution Variant

We solve for the probability by doing $\frac{1-(Probability of Equality)}{2}$.

We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to $(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...$

The summation of this expression is equal to \[\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1\]. Using the geometric sum formula, we obtain the summation of this expression to be $\frac{1}{\frac{3}{4}}-1$ or $\frac{1}{3}$.

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)

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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