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

## 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 (Problems • Answer Key • Resources) Preceded by1980 AHSME Followed by1982 AHSME 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS