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

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The digit <math>7</math> can be either the tens digit (<math>70, 71, \dots, 79</math>: <math>10</math> possibilities), or the ones digit (<math>17, 27, \dots, 97</math>: <math>9</math> possibilities), but we counted the number <math>77</math> twice. This means that out of the <math>90</math> two-digit numbers, <math>10+9-1=18</math> have at least one digit equal to <math>7</math>. Therefore the probability is <math>\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}</math>.
 
The digit <math>7</math> can be either the tens digit (<math>70, 71, \dots, 79</math>: <math>10</math> possibilities), or the ones digit (<math>17, 27, \dots, 97</math>: <math>9</math> possibilities), but we counted the number <math>77</math> twice. This means that out of the <math>90</math> two-digit numbers, <math>10+9-1=18</math> have at least one digit equal to <math>7</math>. Therefore the probability is <math>\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}</math>.
  
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By complimentary counting, we count the numbers that do not contain a <math>7</math>, then subtract from the total. There is a <math>\frac{8}{9}\cdot\frac{9}{10}</math> probability of choosing a number that does NOT contain a <math>7</math>. Subtract this from <math>1</math> and simplify yields <math>1 - \frac{8}{9}\cdot\frac{9}{10} = \frac{90}{90} - \frac{72}{90} = \frac{18}{90} = \frac{1}{5}</math>.
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== Solution 2 ==
 +
By complementary counting, we count the numbers that do not contain a <math>7</math>, then subtract from the total. There is a <math>\frac{8}{9}\cdot\frac{9}{10}</math> probability of choosing a number that does NOT contain a <math>7</math>. Subtract this from <math>1</math> and simplify yields <math>1 - \frac{8}{9}\cdot\frac{9}{10} = \frac{90}{90} - \frac{72}{90} = \frac{18}{90} = \frac{1}{5}</math>.
 +
 
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== Video Solution 1==
 +
https://youtu.be/s1771tqX32k
 +
 
 +
~Education, the Study of Everything
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2004|ab=B|num-b=3|num-a=5}}
 
{{AMC12 box|year=2004|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 18:22, 22 October 2022

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 $1 - \frac{8}{9}\cdot\frac{9}{10} = \frac{90}{90} - \frac{72}{90} = \frac{18}{90} = \frac{1}{5}$.

Video Solution 1

https://youtu.be/s1771tqX32k

~Education, the Study of Everything

See Also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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