2015 AMC 10B Problem 11

The available digits are $2,3,5,7$. There are $4$ one-digit numbers and $4^2=16$ two-digit numbers, for $20$ total choices. All $4$ one-digit choices are prime. A two-digit prime cannot end in $2$ or $5$, so checking endings $3$ and $7$ gives the two-digit primes $23,37,53,73$. Thus $8$ of the $20$ choices are prime, and the probability is $\frac{8}{20}=\frac25$. Thus, the correct answer is B.