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

(Solution: alternate solution)
(Solution 2 (variant))
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Suppose the green ball goes in bin <math>i</math>, for some <math>i \ge 1</math>. The probability of this occurring is <math>\frac{1}{2^i}</math>. Given this occurs, the probability that the red ball goes in a higher-numbered bin is <math>\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}</math>. Thus the probability that the green ball goes in bin <math>i</math>, and the red ball goes in a bin greater than <math>i</math>, is <math>\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}</math>. Summing from <math>i=1</math> to infinity, we get
 
Suppose the green ball goes in bin <math>i</math>, for some <math>i \ge 1</math>. The probability of this occurring is <math>\frac{1}{2^i}</math>. Given this occurs, the probability that the red ball goes in a higher-numbered bin is <math>\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}</math>. Thus the probability that the green ball goes in bin <math>i</math>, and the red ball goes in a bin greater than <math>i</math>, is <math>\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}</math>. Summing from <math>i=1</math> to infinity, we get
  
<cmath> \sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath>
+
<cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath>
 +
 
 +
-scrabbler94
  
 
==See Also==
 
==See Also==

Revision as of 14:47, 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 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

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

-scrabbler94

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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