Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AIME II Problems/Problem 6"

(→‎See also)
(→‎Solution)
Line 3: Line 3:
 
Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability <math>\frac{2}{3}</math> and each of the other five sides has probability <math>\frac{1}{15}</math>. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.  
 
Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability <math>\frac{2}{3}</math> and each of the other five sides has probability <math>\frac{1}{15}</math>. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.  
  
−
==Solution==
+
==Solution 1==
  
 
The probability that he rolls a six twice when using the fair die is <math>\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}</math>.  The probability that he rolls a six twice using the biased die is <math>\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}</math>.  Given that Charles rolled two sixes, we can see that it is <math>16</math> times more likely that he chose the second die.  Therefore the probability that he is using the fair die is <math>\frac{1}{17}</math>, and the probability that he is using the biased die is <math>\frac{16}{17}</math>.  The probability of rolling a third six is
 
The probability that he rolls a six twice when using the fair die is <math>\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}</math>.  The probability that he rolls a six twice using the biased die is <math>\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}</math>.  Given that Charles rolled two sixes, we can see that it is <math>16</math> times more likely that he chose the second die.  Therefore the probability that he is using the fair die is <math>\frac{1}{17}</math>, and the probability that he is using the biased die is <math>\frac{16}{17}</math>.  The probability of rolling a third six is
Line 9: Line 9:
 
<cmath>\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}</cmath>
 
<cmath>\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}</cmath>
 
Therefore, our desired <math>p+q</math> is <math>65+102= \boxed{167}</math>
 
Therefore, our desired <math>p+q</math> is <math>65+102= \boxed{167}</math>
 +
 +
==Solution 2==
 +
This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)!
 +
The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. This is thus
 +
<math>\frac{\frac{1}{2}\frac{2}{3}^3+\frac{1}{2}\frac{1}{6}^3}{\frac{1}{2}\frac{2}{3}^2+\frac{1}{2}\frac{1}{6}^2}= \frac{65}{102}</math>. The answer is therefore <math>65+102= \boxed 167</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:36, 13 June 2021

Problem

Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution 1

The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$. The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$. Given that Charles rolled two sixes, we can see that it is $16$ times more likely that he chose the second die. Therefore the probability that he is using the fair die is $\frac{1}{17}$, and the probability that he is using the biased die is $\frac{16}{17}$. The probability of rolling a third six is

\[\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}\] Therefore, our desired $p+q$ is $65+102= \boxed{167}$

Solution 2

This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)! The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. This is thus $\frac{\frac{1}{2}\frac{2}{3}^3+\frac{1}{2}\frac{1}{6}^3}{\frac{1}{2}\frac{2}{3}^2+\frac{1}{2}\frac{1}{6}^2}= \frac{65}{102}$. The answer is therefore $65+102= \boxed 167$.

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_6&oldid=155897"