Difference between revisions of "1985 AJHSME Problems/Problem 22"
5849206328x (talk | contribs) (New page: ==Problem== Assume every 7-digit whole number is a possible telephone number except those that begin with <math>0</math> or <math>1</math>. What fraction of telephone numbers begin with ...) |
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==Problem== | ==Problem== | ||
| − | Assume every 7-digit whole number is a possible telephone number except those that begin with <math>0</math> or <math>1</math>. What fraction of telephone numbers begin with <math>9</math> and end with <math>0</math>? | + | Assume every 7-digit [[whole number]] is a possible telephone number except those that begin with <math>0</math> or <math>1</math>. What [[fraction]] of telephone numbers begin with <math>9</math> and end with <math>0</math>? |
<math>\text{(A)}\ \frac{1}{63} \qquad \text{(B)}\ \frac{1}{80} \qquad \text{(C)}\ \frac{1}{81} \qquad \text{(D)}\ \frac{1}{90} \qquad \text{(E)}\ \frac{1}{100}</math> | <math>\text{(A)}\ \frac{1}{63} \qquad \text{(B)}\ \frac{1}{80} \qquad \text{(C)}\ \frac{1}{81} \qquad \text{(D)}\ \frac{1}{90} \qquad \text{(E)}\ \frac{1}{100}</math> | ||
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==Solution== | ==Solution== | ||
| − | An equivalent problem is finding the probability that a randomly selected telephone number begins with <math>9</math> and ends with <math>0</math>. | + | An equivalent problem is finding the [[probability]] that a randomly selected telephone number begins with <math>9</math> and ends with <math>0</math>. |
There are <math>10-2=8</math> possibilities for the first digit in total, and only <math>1</math> that works, so the probability the number begins with <math>9</math> is <math>\frac{1}{8}</math> | There are <math>10-2=8</math> possibilities for the first digit in total, and only <math>1</math> that works, so the probability the number begins with <math>9</math> is <math>\frac{1}{8}</math> | ||
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There are <math>10</math> possibilities for the last digit, and only <math>1</math> that works <math>(0)</math>, so the probability the number ends with <math>0</math> is <math>\frac{1}{10}</math> | There are <math>10</math> possibilities for the last digit, and only <math>1</math> that works <math>(0)</math>, so the probability the number ends with <math>0</math> is <math>\frac{1}{10}</math> | ||
| − | Since these are independent events, the probability both happens is <cmath>\frac{1}{8}\cdot \frac{1}{10}=\frac{1}{80}</cmath> | + | Since these are [[Independent event|independent events]], the probability both happens is <cmath>\frac{1}{8}\cdot \frac{1}{10}=\frac{1}{80}</cmath> |
<math>\boxed{\text{B}}</math> | <math>\boxed{\text{B}}</math> | ||
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==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1985|num-b=21|num-a=23}} |
| + | [[Category:Introductory Combinatorics Problems]] | ||
| + | [[Category:Probability Problems]] | ||
Revision as of 19:03, 17 May 2009
Problem
Assume every 7-digit whole number is a possible telephone number except those that begin with 
or 
. What fraction of telephone numbers begin with 
and end with ?
Note: All telephone numbers are 7-digit whole numbers.
Solution
An equivalent problem is finding the probability that a randomly selected telephone number begins with and ends with .
There are 
possibilities for the first digit in total, and only that works, so the probability the number begins with is 
There are 
possibilities for the last digit, and only that works 
, so the probability the number ends with is 
Since these are independent events, the probability both happens is ![\[\frac{1}{8}\cdot \frac{1}{10}=\frac{1}{80}\]](http://latex.artofproblemsolving.com/a/4/5/a456f704d11b2cb826e1964b6c0c52649681eb78.png)
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
