Difference between revisions of "2015 AMC 10B Problems/Problem 11"
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==Solution== | ==Solution== | ||
The one digit prime numbers are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>. So there are a total of <math>4\cdot4=16</math> ways to choose a two digit number with both digits as primes and <math>4</math> ways to choose a one digit prime, for a total of <math>4+16=20</math> ways. Out of these <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>23</math>, <math>37</math>, <math>53</math>, and <math>73</math> are prime. Thus the probability is <math>\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}</math>. | The one digit prime numbers are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>. So there are a total of <math>4\cdot4=16</math> ways to choose a two digit number with both digits as primes and <math>4</math> ways to choose a one digit prime, for a total of <math>4+16=20</math> ways. Out of these <math>2</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>23</math>, <math>37</math>, <math>53</math>, and <math>73</math> are prime. Thus the probability is <math>\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}</math>. | ||
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| + | ==Solution 2== | ||
| + | We count the number of primes. We get 25 primes. The only answer choice that has a denominator of 25 (or simplified) is B. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 19:14, 25 September 2021
Problem
Among the positive integers less than 
, each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?
Solution
The one digit prime numbers are 
, 
, 
, and 
. So there are a total of 
ways to choose a two digit number with both digits as primes and 
ways to choose a one digit prime, for a total of 
ways. Out of these , , , , 
, 
, 
, and 
are prime. Thus the probability is 
.
Solution 2
We count the number of primes. We get 25 primes. The only answer choice that has a denominator of 25 (or simplified) is B.
Video Solution
~savannahsolver
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
