1981 AMC 12 Problem 26

Attempt Problem 26 of the 1981 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1981 AMC 12 solutions, or check the answer key.

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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 16,\frac16, independent of the outcome of any other toss.)

13\frac13

29\frac29

518\frac5{18}

2591\frac{25}{91}

3691\frac{36}{91}

Answer: D
Concepts:dice (probability)geometric distributionindependent events
Difficulty rating: 1680
Small Hint:

For Carol to win in a given round, Alice and Bob must fail before Carol succeeds

Big Hint:

If all three fail, the process restarts with the same probabilities

Solution:

Carol wins in the first round with probability (56)2(16)=25216.(\frac{5}{6})^2(\frac{1}{6})=\frac{25}{216}. A complete round with no six has probability (56)3=125216.(\frac{5}{6})^3=\frac{125}{216}. Therefore the desired geometric series is 252161125216=2591. \frac{\frac{25}{216}}{1-\frac{125}{216}} =\frac{25}{91}.

Therefore, the correct answer is D.

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