A bag contains 13 cubes numbered 1 to 13. Two cubes are drawn sequentially without replacement. What is the probability that both cubes are odd numbered?

There are 7 odd cubes out of 13 total cubes.
$$
P(\text{1st cube is odd}) = \frac{7}{13}
$$

After drawing one odd cube, there are 6 odd cubes left and a total of 12 cubes remaining.
$$
P(\text{2nd cube is odd} | \text{1st cube is odd}) = \frac{6}{12} = \frac{1}{2}
$$

$$
P(\text{both are odd}) = P(\text{1st odd}) \times P(\text{2nd odd} | \text{1st odd})
$$
$$
P(\text{both are odd}) = \frac{7}{13} \times \frac{1}{2}
$$
$$
P(\text{both are odd}) = \frac{7}{26}
$$
The probability that both cubes are odd numbered is $\boxed{\frac{7}{26}}$.

$$
P(\text{1st Even}) = \frac{6}{13}
$$
After drawing one even cube, there are 7 odd cubes left and 12 total cubes remaining.
$$
P(\text{2nd Odd} | \text{1st Even}) = \frac{7}{12}
$$
$$
P(\text{1st Even, 2nd Odd}) = \frac{6}{13} \times \frac{7}{12} = \frac{1}{13} \times \frac{7}{2} = \frac{7}{26}
$$

$$
P(\text{1st Odd}) = \frac{7}{13}
$$
After drawing one odd cube, there are 6 even cubes left and 12 total cubes remaining.
$$
P(\text{2nd Even} | \text{1st Odd}) = \frac{6}{12} = \frac{1}{2}
$$
$$
P(\text{1st Odd, 2nd Even}) = \frac{7}{13} \times \frac{1}{2} = \frac{7}{26}
$$

$$
P(\text{only one is even}) = P(\text{1st Even, 2nd Odd}) + P(\text{1st Odd, 2nd Even})
$$
$$
P(\text{only one is even}) = \frac{7}{26} + \frac{7}{26}
$$
$$
P(\text{only one is even}) = \frac{14}{26} = \frac{7}{13}
$$
The probability that only one cube is even numbered is $\boxed{\frac{7}{13}}$.