2015 AMC 10B Problem 11

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

The sample space is tiny once you notice that only four digits are allowed, so the problem is really a primality check on sixteen numbers. Cutting the work in half by discarding numbers ending in $2$ or $5$ is the same last-digit reasoning that underlies every divisibility rule; then a digit-sum check for $3$ handles most of the rest.

• Forgetting that 1 is not prime (which would add 1, 11, 13, ... to the pool) or miscounting the two-digit primes, e.g. calling 57 = 3 * 19 prime.
• Using $99$ or $100$ as the denominator (choice (A) is the bait) instead of the $20$ numbers that actually satisfy the digit condition.