# 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 1

~Education, the Study of Everything

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 