2015 AMC 10B Problem 11

1. Which digits are allowed? The prime digits are $2, 3, 5, 7$. (The digit $1$ is not prime.)
2. Count all the numbers. One-digit numbers: $2, 3, 5, 7$, which is $4$ numbers. Two-digit numbers: $4$ choices for the first digit and $4$ for the second, which is $4 \cdot 4 = 16$ numbers. In total there are $4 + 16 = 20$ numbers.
3. One-digit primes. All four of $2, 3, 5, 7$ are prime.
4. Two-digit primes. A two-digit number ending in $2$ is even, and one ending in $5$ is a multiple of $5$, so neither can be prime. That leaves the eight numbers ending in $3$ or $7$:

Four of them are prime: $23, 37, 53, 73$.
5. Probability. There are $4 + 4 = 8$ primes out of $20$ numbers, so the probability is $\dfrac{8}{20} = \dfrac25$.

The answer is $\boxed{\textbf{(B)}\ \frac{2}{5}}$.

Once you see that only four digits are allowed, the list of candidates is small enough to check by hand. The last digit throws out half the list at once. For the rest, the digit-sum test for $3$ catches $33$, $27$ and $57$, and $77$ is a multiple of $7$.

The prime digits are $2, 3, 5, 7$. The one-digit candidates are these four numbers, and the two-digit candidates have $4$ choices for each digit, giving $4 \cdot 4 = 16$ numbers. The pool has $4 + 16 = 20$ numbers.

All four one-digit candidates are prime.

Among the two-digit candidates, any number ending in $2$ or $5$ is composite, so only those ending in $3$ or $7$ can be prime: $23, 33, 53, 73, 27, 37, 57, 77$. Checking each: $23$ prime, $33 = 3 \cdot 11$, $53$ prime, $73$ prime, $27 = 3^3$, $37$ prime, $57 = 3 \cdot 19$, $77 = 7 \cdot 11$. Exactly four are prime.

So $4 + 4 = 8$ of the $20$ numbers are prime, and the probability is $\frac{8}{20} = \frac{2}{5}$.

The answer is $\boxed{\textbf{(B)}\ \frac{2}{5}}$.