一个袋子里装有四个红球 and 八个蓝球，其中两个红球和六个蓝球上印有数字“0”。其余的球上印有数字“1”。从袋中无放回地随机抽取三个球。

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:
$P(\text{at least one of each}) = \frac{8}{11}$