Difference between revisions of "2004 AMC 12B Problems/Problem 4"
(New page: == Problem == An integer <math>x</math>, with <math>10\leq x\leq 99</math>, is to be chosen. If all choices are equally likely, what is the probability that at least one digit of <math>x<...) |
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== Solution == | == Solution == | ||
| − | The digit 7 can be either the tens digit (<math>70, 71, \dots, 79</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>. |
== 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}} | ||
Revision as of 09:08, 6 January 2009
Problem
An integer 
, with 
, is to be chosen. If all choices are equally likely, what is the probability that at least one digit of is a 7?
Solution
The digit 
can be either the tens digit (
: 
possibilities), or the ones digit (
: 
possibilities), but we counted the number 
twice. This means that out of the 
two-digit numbers, 
have at least one digit equal to . Therefore the probability is 
.
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 | |
