Difference between revisions of "2015 AMC 10B Problems/Problem 11"
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==Problem== | ==Problem== | ||
| − | Among the positive integers less than 100, each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime? | + | Among the positive integers less than <math>100</math>, each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime? |
<math>\textbf{(A)} \dfrac{8}{99}\qquad \textbf{(B)} \dfrac{2}{5}\qquad \textbf{(C)} \dfrac{9}{20}\qquad \textbf{(D)} \dfrac{1}{2}\qquad \textbf{(E)} \dfrac{9}{16}</math> | <math>\textbf{(A)} \dfrac{8}{99}\qquad \textbf{(B)} \dfrac{2}{5}\qquad \textbf{(C)} \dfrac{9}{20}\qquad \textbf{(D)} \dfrac{1}{2}\qquad \textbf{(E)} \dfrac{9}{16}</math> | ||
Revision as of 16:41, 13 July 2020
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 4 ways to choose a one digit prime, for a total of 
ways. Out of these , , , , 
, 
, 
, and 
are prime. Thus the probability is 
.
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. 
