Answer: 2/3

Find the probability that neither student is a junior, then subtract it from $1$. The probability that both are seniors is $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$. So the probability that at least one is a junior is $1 - \frac{1}{3} = \frac{2}{3}$.

Using $\frac{6}{10} \times \frac{6}{10}$ for both seniors treats the selection as if it were with replacement and gives $0.64$ instead of $\frac{2}{3}$.