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

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<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>
  
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==Solution==
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==Solution 1==
 
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 (Listing)==
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Since the only primes digits are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>, it doesn't seem too hard to list all of the numbers out.
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<math>2</math>- Prime;
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<math>3</math>- Prime;
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<math>5</math>- Prime;
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<math>7</math>- Prime;
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<math>22</math>- Composite;
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<math>23</math>- Prime;
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<math>25</math>- Composite;
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<math>27</math>- Composite;
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<math>32</math>- Composite;
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<math>33</math>- Composite;
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<math>35</math>- Composite;
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<math>37</math>- Prime;
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<math>52</math>- Composite;
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<math>53</math>- Prime;
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<math>55</math>- Composite;
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<math>57</math>- Composite;
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<math>72</math>- Composite;
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<math>73</math>- Prime;
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<math>75</math>- Composite;
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<math>77</math>- Composite.
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Counting it out, there are <math>20</math> cases and <math>8</math> of these are prime. So the answer is <math>\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}</math>.
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~JH. L
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:25, 16 June 2022

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 1

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 $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$.

Solution 2 (Listing)

Since the only primes digits are $2$, $3$, $5$, and $7$, it doesn't seem too hard to list all of the numbers out. $2$- Prime; $3$- Prime; $5$- Prime; $7$- Prime; $22$- Composite; $23$- Prime; $25$- Composite; $27$- Composite; $32$- Composite; $33$- Composite; $35$- Composite; $37$- Prime; $52$- Composite; $53$- Prime; $55$- Composite; $57$- Composite; $72$- Composite; $73$- Prime; $75$- Composite; $77$- Composite. Counting it out, there are $20$ cases and $8$ of these are prime. So the answer is $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$. ~JH. L

Video Solution

https://youtu.be/cL9wo9kcOGg

~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

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