jmc

There are 16 multiples of 3 between 1 and 50 ($1\cdot 3$ through $16\cdot 3$), and $50-16=34$ numbers which are not multiples of 3. The probability that neither of the numbers Ben chooses is a multiple of 3 is ${\left(\frac{34}{50}\right)}^2=\frac{1156}{2500}$. Therefore, the probability that at least one of the numbers that Ben chooses is a multiple of 3 is $1-\frac{1156}{2500}=\frac{1344}{2500}=\boxed{\frac{336}{625}}$.