Difference between revisions of "1981 AHSME Problems/Problem 26"
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− | == Problem == | + | ==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 <math> \dfrac{1}{6}</math>, independent of the outcome of any other toss.) | ||
− | + | <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) } \ | ||
− | == Solution 1 == | + | == Solution 1 (Infinite Series)== |
− | + | 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> | + | 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 , independent of the outcome of any other toss.)
Solution 1 (Infinite Series)
For Carol to win on her roll, the first
rolls must not be
and the last roll must be
. Therefore, the probability that Carol wins on her
roll is
. Taking the sum of this expression over all positive integers
, her total probability of winning must be
. Since this is a geometric series with first term
and common ratio
, Carol's probability of winning is
.
Solution 2 (Probability States)
Let ,
, and
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
will be the answer.
Then ,
, and
. Substituting into the first equation shows that
. Solving the equation
gives
.
-j314andrews
See also
1981 AHSME (Problems • Answer Key • Resources) | ||
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.