2004 AMC 12B Problem 4

Attempt Problem 4 of the 2004 AMC 12B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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}

Count two-digit numbers with a

7

in the units place and with a

7

in the tens place

77

is counted in both groups, so subtract it once

There are

90

integers from

10

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.

Problem 4 in other years