2018 AMC 10B Problem 5

Count the complement. The set has $2^8 = 256$ subsets total. A subset avoids every prime exactly when it sticks to the non-primes $\{4, 6, 8, 9\},$ and there are $2^4 = 16$ of those. So $256 - 16 = 240$ subsets contain at least one prime. Thus, D is the correct answer.