2004 AMC 12B Problem 4

Below is the professionally curated solution for Problem 4 of the 2004 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12B solutions, or check the answer key.

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Concepts:inclusion-exclusion digits basic probability

Difficulty rating: 1100

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

$\dfrac{1}{9}$

$\dfrac{1}{5}$

$\dfrac{19}{90}$

$\dfrac{2}{9}$

$\dfrac{1}{3}$

Solution:

There are $90$ integers from $10$ to $99.$ Ten have a units digit $7,$ and nine have a tens digit $7.$ Since $77$ is counted twice, there are $10 + 9 - 1 = 18$ with at least one $7.$ The probability is $\dfrac{18}{90} = \dfrac{1}{5}.$

Thus, the correct answer is B.

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