Difference between revisions of "1981 AHSME Problems/Problem 26"

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== Problem ==
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==Problem 26==
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Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is <math> \dfrac{1}{6}</math>, independent of the outcome of any other toss.)
  
Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is <math>\frac{1}{6}</math>, independent of the outcome of any other toss.)
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<math> \textbf{(A)}\ \dfrac{1}{3}\qquad \textbf{(B)}\ \dfrac{2}{9}\qquad \textbf{(C)}\ \dfrac{5}{18}\qquad \textbf{(D)}\ \dfrac{25}{91}\qquad \textbf{(E)}\ \dfrac{36}{91}</math>
<math>\textbf{(A) } \frac{1}{3} \textbf{(B) } \frac{2}{9} \textbf{(C) } \frac{5}{18} \textbf{(D) } \frac{25}{91} \textbf{(E) } \frac{36}{91}</math>
 
  
== Solution 1 ==
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== Solution 1 (Infinite Series)==
  
The probability that Carol wins during the first cycle through is <math>\frac{5}{6}*\frac{5}{6}*\frac{1}{6}</math>, and the probability that Carol wins on the second cycle through is <math>\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}</math>. It is clear that this is an infinite geometric sequence, and we must find the sum of it in order to find the answer to this question. Thus we set up the equation: <math>\frac{\frac{25}{216}}{1-\frac{125}{216}}</math>, or <math>\frac{\frac{25}{216}}{\frac{91}{216}}</math>, which simplifies into <math>\boxed{\textbf{(D) } \frac{25}{91}}</math>
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For Carol to win on her <math>n^{th}</math> roll, the first <math>3n-1</math> rolls must not be <math>6</math> and the last roll must be <math>6</math>.  Therefore, the probability that Carol wins on her <math>n^{th}</math> roll is <math>\frac{1}{6} \cdot \left(\frac{5}{6}\right)^{3n-1} = \frac{1}{5} \cdot \left(\frac{125}{216}\right)^{n}</math>.  Taking the sum of this expression over all positive integers <math>n</math>, her total probability of winning must be <math>\sum_{n=1}^{\infty} \frac{1}{5} \cdot \left(\frac{125}{216}\right)^{n}</math>.  Since this is a geometric series with first term <math>\frac{25}{216}</math> and common ratio <math>\frac{125}{216}</math>, Carol's probability of winning is  <math>\frac{\frac{25}{216}}{1-\frac{125}{216}}=\frac{\frac{25}{216}}{\frac{91}{216}}=\frac{25}{91}</math>. <math>\fbox{(D)}</math>
  
 
== Solution 2 (Probability States) ==
 
== Solution 2 (Probability States) ==
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Let <math>a</math>, <math>b</math>, and <math>c</math> be Carol's probabilities of winning, depending on whether it is Alice's, Bob's or Carol's turn to roll the die next, respectively.  Since it is initially Alice's turn, the value of <math>a</math> will be the answer.
 
Let <math>a</math>, <math>b</math>, and <math>c</math> be Carol's probabilities of winning, depending on whether it is Alice's, Bob's or Carol's turn to roll the die next, respectively.  Since it is initially Alice's turn, the value of <math>a</math> will be the answer.
  
Then <math>a = \frac{5}{6}b</math>, <math>b = \frac{5}{6}c</math>, and <math>c = \frac{1}{6} + \frac{5}{6}a</math>.  Substituting into the first equation shows that <math>a = \frac{5}{6}b = \frac{25}{36}c = \frac{25}{36}\left(\frac{1}{6} + \frac{5}{6}a\right) = \frac{25}{216} + \frac{125}{216}a</math>. Solving the equation <math>a = \frac{25}{216} + \frac{125}{216}a</math> gives <math>a = \frac{25}{91}</math> <math>\fbox{(D)}</math>  
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Then <math>a = \frac{5}{6}b</math>, <math>b = \frac{5}{6}c</math>, and <math>c = \frac{1}{6} + \frac{5}{6}a</math>.  Substituting into the first equation shows that <math>a = \frac{5}{6}b = \frac{25}{36}c = \frac{25}{36}\left(\frac{1}{6} + \frac{5}{6}a\right) = \frac{25}{216} + \frac{125}{216}a</math>. Solving the equation <math>a = \frac{25}{216} + \frac{125}{216}a</math> gives <math>a = \frac{25}{91}</math>. <math>\fbox{(D)}</math>  
  
 
-j314andrews
 
-j314andrews

Latest revision as of 13:48, 29 June 2025

Problem 26

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\dfrac{1}{6}$, independent of the outcome of any other toss.)

$\textbf{(A)}\ \dfrac{1}{3}\qquad \textbf{(B)}\ \dfrac{2}{9}\qquad \textbf{(C)}\ \dfrac{5}{18}\qquad \textbf{(D)}\ \dfrac{25}{91}\qquad \textbf{(E)}\ \dfrac{36}{91}$

Solution 1 (Infinite Series)

For Carol to win on her $n^{th}$ roll, the first $3n-1$ rolls must not be $6$ and the last roll must be $6$. Therefore, the probability that Carol wins on her $n^{th}$ roll is $\frac{1}{6} \cdot \left(\frac{5}{6}\right)^{3n-1} = \frac{1}{5} \cdot \left(\frac{125}{216}\right)^{n}$. Taking the sum of this expression over all positive integers $n$, her total probability of winning must be $\sum_{n=1}^{\infty} \frac{1}{5} \cdot \left(\frac{125}{216}\right)^{n}$. Since this is a geometric series with first term $\frac{25}{216}$ and common ratio $\frac{125}{216}$, Carol's probability of winning is $\frac{\frac{25}{216}}{1-\frac{125}{216}}=\frac{\frac{25}{216}}{\frac{91}{216}}=\frac{25}{91}$. $\fbox{(D)}$

Solution 2 (Probability States)

Let $a$, $b$, and $c$ be Carol's probabilities of winning, depending on whether it is Alice's, Bob's or Carol's turn to roll the die next, respectively. Since it is initially Alice's turn, the value of $a$ will be the answer.

Then $a = \frac{5}{6}b$, $b = \frac{5}{6}c$, and $c = \frac{1}{6} + \frac{5}{6}a$. Substituting into the first equation shows that $a = \frac{5}{6}b = \frac{25}{36}c = \frac{25}{36}\left(\frac{1}{6} + \frac{5}{6}a\right) = \frac{25}{216} + \frac{125}{216}a$. Solving the equation $a = \frac{25}{216} + \frac{125}{216}a$ gives $a = \frac{25}{91}$. $\fbox{(D)}$

-j314andrews

See also

1981 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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 26 27 28 29 30
All AHSME Problems and Solutions

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