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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==
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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 4 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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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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*2- Prime;
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*3- Prime;
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*5- Prime;
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*7- Prime;
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*22- Composite;
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*23- Prime;
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*25- Composite;
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*27- Composite;
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*32- Composite;
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*33- Composite;
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*35- Composite;
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*37- Prime;
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*52- Composite;
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*53- Prime;
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*55- Composite;
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*57- Composite;
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*72- Composite;
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*73- Prime;
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*75- Composite;
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*77- 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
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 +
==Video Solution 1==
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https://youtu.be/RZDFs3qrw7Y
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/cL9wo9kcOGg
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2015|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:13, 2 August 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 1

https://youtu.be/RZDFs3qrw7Y

~Education, the Study of Everything

Video Solution

https://youtu.be/cL9wo9kcOGg

~savannahsolver

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AMC 10 Problems and Solutions

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