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2015 AMC 10B Problems/Problem 11: Difference between revisions

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==Solution 2 (Listing)==
==Solution 2 (Listing)==
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.
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.
<math>2</math>- Prime;
 
<math>3</math>- Prime;
*2- Prime;
<math>5</math>- Prime;
*3- Prime;
<math>7</math>- Prime;
*5- Prime;
<math>22</math>- Composite;
*7- Prime;
<math>23</math>- Prime;
*22- Composite;
<math>25</math>- Composite;
*23- Prime;
<math>27</math>- Composite;
*25- Composite;
<math>32</math>- Composite;
*27- Composite;
<math>33</math>- Composite;
*32- Composite;
<math>35</math>- Composite;
*33- Composite;
<math>37</math>- Prime;
*35- Composite;
<math>52</math>- Composite;
*37- Prime;
<math>53</math>- Prime;
*52- Composite;
<math>55</math>- Composite;
*53- Prime;
<math>57</math>- Composite;
*55- Composite;
<math>72</math>- Composite;
*57- Composite;
<math>73</math>- Prime;
*72- Composite;
<math>75</math>- Composite;
*73- Prime;
<math>77</math>- Composite.
*75- Composite;
*77- Composite.
 
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>.
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>.
~JH. L
~JH. L

Revision as of 22:30, 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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