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

(Solution 6)
(Solution 6)
Line 30: Line 30:
  
 
==Solution 6 ==
 
==Solution 6 ==
Probability that the green ball is in a bin below red for red in bin <math>k</math> is <math>\sum\limits_{i=1}^{k-1}2^{-i}</math>. There's a <math>2^{-k}</math> chance for red to be in that bin. Then, <math>P(Green<Red|Red=k)=\sum\limits_{k=1}^{\infty}2^{-k}\sum\limits_{i=1}^{k-1}2^{-i}</math>. This is  
+
Probability that the green ball is in a bin below red for red in bin <math>k</math> is <math>\sum\limits_{i=1}^{k-1}2^{-i}</math>. There's a <math>2^{-k}</math> chance for red to be in that bin. Then, <math>P(\text{Green below Red}|\text{Red}=k)=\sum\limits_{k=1}^{\infty}2^{-k}\sum\limits_{i=1}^{k-1}2^{-i}</math>. This is  
 
<math>\sum\limits_{i=1}^{\infty}2^{-i}\cdot\frac{1}{2}\frac{1-(1/2)^k-1}{1-1/2}</math> or,
 
<math>\sum\limits_{i=1}^{\infty}2^{-i}\cdot\frac{1}{2}\frac{1-(1/2)^k-1}{1-1/2}</math> or,
 
<cmath>\sum\limits_{i=1}^{\infty}\frac{1}{2^{-i}}-2\sum\limits_{i=1}^\infty\frac{1}{(2^2)^{-i}}</cmath>
 
<cmath>\sum\limits_{i=1}^{\infty}\frac{1}{2^{-i}}-2\sum\limits_{i=1}^\infty\frac{1}{(2^2)^{-i}}</cmath>

Revision as of 23:43, 13 December 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 1

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 $\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}$ (by the geometric series sum formula). Therefore the other two probabilities have to both be $\frac{1-\frac{1}{3}}{2} = \boxed{\textbf{(C) } \frac{1}{3}}$.

Solution 2

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

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

Solution 3

The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}$ by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of $2$ from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}$ (again recognizing a geometric series). We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots$, which, by the geometric series sum formula, is $\boxed{\textbf{(C) } \frac{1}{3}}$.

-fidgetboss_4000

Solution 4 (quick, conceptual)

Define a win as a ball appearing in higher numbered box.

Start from the first box.

There are $4$ possible results in the box: Red, Green, Red and Green, or none, with an equal probability of $\frac{1}{4}$ for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if $p$ is the probability that Red wins, we can write $p = \frac{1}{4} + \frac{1}{4}p$: there is a $\frac{1}{4}$ probability that "Red" wins immediately, a $0$ probability in the cases "Green" or "Red and Green", and in the "None" case (which occurs with $\frac{1}{4}$ probability), we then start again, giving the same probability $p$. Hence, solving the equation, we get $p = \boxed{\textbf{(C) } \frac{1}{3}}$.

Solution 5

Write out the infinite geometric series as $\frac{1}{2}$, $\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \cdots$. To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term $1$, term $3$, etc.), and then sum the remaining terms - this is in fact precisely equivalent to the method of Solution 2. Writing this out as another infinite geometric sequence, we are left with $\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \cdots$. Summing, we get \[\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}\]

Solution 6

Probability that the green ball is in a bin below red for red in bin $k$ is $\sum\limits_{i=1}^{k-1}2^{-i}$. There's a $2^{-k}$ chance for red to be in that bin. Then, $P(\text{Green below Red}|\text{Red}=k)=\sum\limits_{k=1}^{\infty}2^{-k}\sum\limits_{i=1}^{k-1}2^{-i}$. This is $\sum\limits_{i=1}^{\infty}2^{-i}\cdot\frac{1}{2}\frac{1-(1/2)^k-1}{1-1/2}$ or, \[\sum\limits_{i=1}^{\infty}\frac{1}{2^{-i}}-2\sum\limits_{i=1}^\infty\frac{1}{(2^2)^{-i}}\] \[1-2/3 \implies \boxed{D) 1/3}\]

~BJHHar

Video Solution

For those who want a video solution: https://youtu.be/VP7ltu-XEq8

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS