1977 AMC 12 Problem 17

Question

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?

\frac16

\frac19

\frac1{27}

\frac1{54}

\frac7{36}

Answer: B

Concepts:basic probability dice (probability)permutations

Difficulty rating: 1590

Small hint:

List the possible three-element sets of consecutive die values

Big hint:

Each qualifying set has six orderings among the

6^3

ordered outcomes

Solution:

The possible sets are

\{1,2,3\},

\{2,3,4\},

\{3,4,5\},

and

\{4,5,6\}.

Each has

3!=6

orderings, so

24

of the

6^3=216

outcomes work. The probability is

\frac{24}{216}=\frac{1}{9}.

Therefore, the correct answer is B.