2015 AMC 10B Problems/Problem 11

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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 (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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