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Difference between revisions of "2015 AMC 10B Problems/Problem 11"

(Solution 2)
(Solution 2)
 
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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>.
 
==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==

Latest revision as of 15:07, 17 October 2021

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?

$\textbf{(A)} \dfrac{8}{99}\qquad \textbf{(B)} \dfrac{2}{5}\qquad \textbf{(C)} \dfrac{9}{20}\qquad \textbf{(D)} \dfrac{1}{2}\qquad \textbf{(E)} \dfrac{9}{16}$

Solution

The one digit prime numbers are $2$, $3$, $5$, and $7$. So there are a total of $4\cdot4=16$ 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 $4+16=20$ ways. Out of these $2$, $3$, $5$, $7$, $23$, $37$, $53$, and $73$ are prime. Thus the probability is $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$.

Video Solution

https://youtu.be/cL9wo9kcOGg

~savannahsolver

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
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

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