A bag contains 4 red counters and 6 blue counters. 4 counters are drawn from the bag at random, without replacement.

1. Set up the probability using combinations for selecting 4 blue counters from 6, out of a total of 4 counters from 10.

$\mathrm{P}(\text{all 4 blue}) = \frac{^{6}C_{4}}{^{10}C_{4}}$

2. Evaluate the combinations to find the simplified final probability.

$= \frac{15}{210} \\ = \frac{1}{14}$

1. Calculate the probability of drawing exactly 4 blue counters (derived from part i).

$\text{P}(\text{4 blue}) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} = \frac{1}{14}$

2. Calculate the probability of drawing exactly 3 blue counters, considering the different possible arrangements.

A result of exactly 3 blue counters and 1 red counter can occur in 4 possible ways (e.g., BBBR, BBRB, BRBB, RBBB).
$\text{P}(\text{3 blue, 1 red}) = 4 \times \left( \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{4}{7} \right)$
$= 4 \times \frac{2}{21} = \frac{8}{21}$

3. Add the probabilities of these mutually exclusive events to find the total probability of drawing at least 3 blue counters.

$\text{P}(\text{at least 3 blue}) = \text{P}(\text{4 blue}) + \text{P}(\text{3 blue, 1 red})$
$= \frac{1}{14} + \frac{8}{21}$
$= \frac{3}{42} + \frac{16}{42} = \frac{19}{42}$

1. Calculate the probability that all 4 counters drawn are red.

$\text{P}(\text{all 4 are red}) = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7} = \frac{1}{210}$

2. Use the complement rule to find the probability of drawing at least one counter of each colour.

From part (a)(i), we know the probability of drawing all blue counters:
$\text{P}(\text{all 4 are blue}) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} = \frac{15}{210} = \frac{1}{14}$

The event "at least 1 counter of each colour" is the complement of "all 4 counters are the same colour".
$\text{P}(\text{at least 1 of each colour}) = 1 - \text{P}(\text{all 4 are blue}) - \text{P}(\text{all 4 are red})$
$= 1 - \frac{15}{210} - \frac{1}{210}$
$= 1 - \frac{16}{210}$
$= \frac{194}{210} = \frac{97}{105}$

1. Determine the probability of the given condition (at least 1 counter of each colour).

Let $B$ be the event of drawing at least 1 counter of each colour (the complement of drawing all red or all blue).
$\text{P}(B) = 1 - \text{P(4 red)} - \text{P(4 blue)}$
$\text{P}(B) = 1 - \frac{\binom{4}{4}}{\binom{10}{4}} - \frac{\binom{6}{4}}{\binom{10}{4}} = 1 - \frac{1}{210} - \frac{15}{210} = \frac{194}{210} = \frac{97}{105}$

2. Determine the probability of the intersection of the two events.

Let $A$ be the event of drawing at least 3 blue counters.
The intersection $A \cap B$ means drawing at least 3 blue counters AND at least 1 counter of each colour, which is exactly 3 blue counters and 1 red counter.
$\text{P}(A \cap B) = \frac{\binom{6}{3} \times \binom{4}{1}}{\binom{10}{4}} = \frac{20 \times 4}{210} = \frac{80}{210} = \frac{8}{21}$

3. Apply the conditional probability formula.

$\text{P}(A \mid B) = \frac{\text{P}(A \cap B)}{\text{P}(B)} = \frac{8/21}{97/105} = \frac{8}{21} \times \frac{105}{97} = \frac{40}{97}$