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Difference between revisions of "2019 AMC 12B Problems/Problem 13"

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==Solution Variant==
 
==Solution Variant==
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We solve for the probability by doing <math>\frac{1-(Probability of Equality)}{2}</math>.
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We solve for the probability by doing <math>\frac{1-(\text{Probability of Equality})}{2}</math>.
  
 
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 <math>(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...</math>  
 
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 <math>(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...</math>  

Revision as of 16:40, 14 February 2019

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 Variant

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

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)

Solution 3 (infinite geometric series)

The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} = \frac{1}{6}$. The probability that the two balls will go into bins that have a distance of 2 from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} = \frac{1}{12}$. We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + ...$, which converges into $\frac{1}{3}$.


See Also

2019 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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