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Difference between revisions of "1977 AHSME Problems/Problem 17"

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Latest revision as of 12:01, 22 November 2016

Problem 17

Three fair dice are tossed at random (i.e., all faces have the same probability of coming up). What is the probability that the three numbers turned up can be arranged to form an arithmetic progression with common difference one?

$\textbf{(A) }\frac{1}{6}\qquad \textbf{(B) }\frac{1}{9}\qquad \textbf{(C) }\frac{1}{27}\qquad \textbf{(D) }\frac{1}{54}\qquad \textbf{(E) }\frac{7}{36}$


Solution

Solution by e_power_pi_times_i


The denominator of the probability is $6^3 = 216$. We are asked to find the probability that the three numbers are consecutive. The three numbers can be arranged in $6$ ways, and there are $4$ triplets of consecutive numbers from $1-6$. The probability is $\dfrac{6\cdot4}{216} = \dfrac{24}{216} = \boxed{\text{(B) }\dfrac{1}{9}}$.

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