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

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 $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$.

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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