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

(Solution)
(Solution 2 (Listing))
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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions

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