A bag contains four red and eight blue balls of which two of the red balls and six of the blue balls have the number "0" printed on them. The remaining balls have the number "1" printed on them. Three balls are randomly drawn from the bag without replacement.

1. Calculate the probability of the complementary event (drawing no blue balls, meaning all three balls are red).

$\text{Total number of balls} = 4 \text{ (red)} + 8 \text{ (blue)} = 12$

$\text{P(all 3 red)} = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{1}{55}$

2. Use the complement rule to find the probability of drawing at least one blue ball.

$P(\text{at least 1 blue}) = 1 - P(\text{all 3 red})$

$P(\text{at least 1 blue}) = 1 - \frac{1}{55} = \frac{54}{55}$

1. Calculate the probability that all 3 balls drawn are blue.

$P(\text{all 3 blue}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{14}{55}$

2. Subtract the probability of drawing all blue balls from the probability of drawing at least one blue ball.

The event 'at least one ball of each colour' is satisfied if we draw at least one blue ball, but we must exclude the outcome where all three balls are blue.
Using the result from part (i):
$P(\text{at least one of each}) = P(\text{at least one blue}) - P(\text{all 3 blue})$
$P(\text{at least one of each}) = \frac{54}{55} - \frac{14}{55} = \frac{40}{55}$
Simplifying the fraction:
$P(\text{at least one of each}) = \frac{8}{11}$