2015 AMC 10B Problems/Problem 11

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

2015 AMC 10B (Problems • Answer Key • Resources)
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2015 AMC 10B Problems/Problem 11

Contents

Problem

Solution 1

Solution 2 (Listing)

Video Solution

See Also