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2004 AMC 12B Problem 4

Problem 4 of 25EasierNumber TheoryCounting & Probability

An integer x,x, with 10≤x≤99,10 \le x \le 99, is to be chosen. If all choices are equally likely, what is the probability that at least one digit of xx is a 7?7?

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Solution

There are 9090 integers from 1010 to 99.99. Ten have a units digit 7,7, and nine have a tens digit 7.7. Since 7777 is counted twice, there are 10+9−1=1810 + 9 - 1 = 18 with at least one 7.7. The probability is 1890=15.\dfrac{18}{90} = \dfrac{1}{5}. Thus, the correct answer is B.

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Concepts: inclusion-exclusion · digits · basic probability

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.