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

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== Problem ==
 
== Problem ==
  
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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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) } \frac{1}{3} \textbf{(B) } \frac{2}{9} \textbf{(C) } \frac{5}{18} \textbf{(D) } \frac{25}{91} \textbf{(E) } \frac{36}{91}</math>
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== Solution ==
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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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== See also ==
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* [[AMC 12 Problems and Solutions]]
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* [[Mathematics competition resources]]
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{{AHSME box|year=1981|before=[[1980 AHSME]]|after=[[1982 AHSME]]}} 
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{{MAA Notice}}

Latest revision as of 00:38, 17 January 2021

Problem

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 $\frac{1}{6}$, independent of the outcome of any other toss.) $\textbf{(A) } \frac{1}{3} \textbf{(B) } \frac{2}{9} \textbf{(C) } \frac{5}{18} \textbf{(D) } \frac{25}{91} \textbf{(E) } \frac{36}{91}$

Solution

The probability that Carol wins during the first cycle through is $\frac{5}{6}*\frac{5}{6}*\frac{1}{6}$, and the probability that Carol wins on the second cycle through is $\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}$. 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: $\frac{\frac{25}{216}}{1-\frac{125}{216}}$, or $\frac{\frac{25}{216}}{\frac{91}{216}}$, which simplifies into $\boxed{\textbf{(D) } \frac{25}{91}}$

See also

1981 AHSME (ProblemsAnswer KeyResources)
Preceded by
1980 AHSME
Followed by
1982 AHSME
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All AHSME Problems and Solutions


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