There are 3 even numbers (2, 4, 6) and 2 odd numbers (1, 7) in the box.

The total number of ways to choose 3 discs from 5 is given by \(\binom{5}{3} = 10\).

To find the number of favorable outcomes (2 even and 1 odd), we can choose 2 even numbers from 3 and 1 odd number from 2:

\(\binom{3}{2} \times \binom{2}{1} = 3 \times 2 = 6\).

Thus, the probability is:

\(\frac{6}{10} = 0.6\).