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AMC 12 Mock Exam 4 Problem 9

Problem 9 of 25

A teacher writes the numbers 11 through 1010 on ten cards, one number per card, shuffles the cards, and deals three of them to Ana. Ana wins a prize if she can arrange her three numbers in some order so that they form an arithmetic progression (for example, 22, 55, 88). What is the probability that Ana wins a prize?

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Solution

A 33-element subset forms an arithmetic progression exactly when it is {m−d,m,m+d}\{m-d,m,m+d\} for some d≥1d\ge1. With 1≤m−d1\le m-d and m+d≤10m+d\le10, there are 10−2d10-2d choices of mm, for d=1,2,3,4d=1,2,3,4: 8+6+4+2=208+6+4+2=20 subsets. There are (103)=120\binom{10}3=120 subsets in all, so the probability is 20120=16\frac{20}{120}=\boxed{\tfrac16}.

Tagged: probability: arithmetic progressions in a random subset