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

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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #17]] and [[2019 AMC 12B Problems|2019 AMC 12B #13]]}}
 
==Problem==
 
==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 <math>k</math> is <math>2^{-k}</math> for <math>k = 1,2,3....</math>  What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
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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 <math>k</math> is <math>2^{-k}</math> for <math>k = 1,2,3....</math>  What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?<br>
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<math>\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}</math>
  
 
==Solution==
 
==Solution==
 
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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 <math>\sum_{k=1}^{\infty}2^{-2k}</math>. The sum is equal to <math>\frac{1}{3}</math>. Therefore the other two probabilities have to both be <math>\textbf{(C) } \frac{1}{3}</math>.<br>
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<math>Q.E.D\blacksquare</math><br>
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Solution by [[User:a1b2|a1b2]]
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2019|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}
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{{MAA Notice}}

Revision as of 13:52, 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

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

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