# Difference between revisions of "2004 AMC 12B Problems/Problem 4"

## Problem

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

$(\mathrm {A}) \dfrac{1}{9} \qquad (\mathrm {B}) \dfrac{1}{5} \qquad (\mathrm {C}) \dfrac{19}{90} \qquad (\mathrm {D}) \dfrac{2}{9} \qquad (\mathrm {E}) \dfrac{1}{3}$

## Solution

The digit $7$ can be either the tens digit ($70, 71, \dots, 79$: $10$ possibilities), or the ones digit ($17, 27, \dots, 97$: $9$ possibilities), but we counted the number $77$ twice. This means that out of the $90$ two-digit numbers, $10+9-1=18$ have at least one digit equal to $7$. Therefore the probability is $\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}$.

## Solution 2

By complementary counting, we count the numbers that do not contain a $7$, then subtract from the total. There is a $\frac{8}{9}\cdot\frac{9}{10}$ probability of choosing a number that does NOT contain a $7$. Subtract this from $1$ and simplify yields $1 - \frac{8}{9}\cdot\frac{9}{10} = \frac{90}{90} - \frac{72}{90} = \frac{18}{90} = \frac{1}{5}$.

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