Use combination method for a

There are $5 \text{ blue} + 4 \text{ green} + 3 \text{ yellow} = 12$ balls in total.
The number of ways to choose 3 balls from 12 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

$$\text{Total ways} = \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$

This can happen in three mutually exclusive ways: all three are blue, or all three are green, or all three are yellow.

• Number of ways to draw 3 blue balls: $\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$
• Number of ways to draw 3 green balls: $\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$
• Number of ways to draw 3 yellow balls: $\binom{3}{3} = \frac{3 \times 2 \times 1}{3 \times 2 \times 1} = 1$ The total number of ways for all three balls to have the same colour is the sum of these possibilities:

$$\text{Favorable ways} = 10 + 4 + 1 = 15$$

$$P(\text{all three same colour}) = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{15}{220} = \frac{3}{44}$$

The probability that all three balls have the same colour is $\boxed{\frac{3}{44}}$.

a) ii. Two balls have the same colour.

This involves choosing 2 balls of one colour and 1 ball from the remaining colours.

• 2 Blue and 1 Green: $\binom{5}{2} \times \binom{4}{1} = \left(\frac{5 \times 4}{2 \times 1}\right) \times 4 = 10 \times 4 = 40$
• 2 Blue and 1 Yellow: $\binom{5}{2} \times \binom{3}{1} = \left(\frac{5 \times 4}{2 \times 1}\right) \times 3 = 10 \times 3 = 30$
• 2 Green and 1 Blue: $\binom{4}{2} \times \binom{5}{1} = \left(\frac{4 \times 3}{2 \times 1}\right) \times 5 = 6 \times 5 = 30$
• 2 Green and 1 Yellow: $\binom{4}{2} \times \binom{3}{1} = \left(\frac{4 \times 3}{2 \times 1}\right) \times 3 = 6 \times 3 = 18$
• 2 Yellow and 1 Blue: $\binom{3}{2} \times \binom{5}{1} = \left(\frac{3 \times 2}{2 \times 1}\right) \times 5 = 3 \times 5 = 15$
• 2 Yellow and 1 Green: $\binom{3}{2} \times \binom{4}{1} = \left(\frac{3 \times 2}{2 \times 1}\right) \times 4 = 3 \times 4 = 12$ The total number of ways for exactly two balls to have the same colour is the sum of these possibilities:

$$\text{Favorable ways} = 40 + 30 + 30 + 18 + 15 + 12 = 145$$

$$P(\text{two balls same colour}) = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{145}{220} = \frac{29}{44}$$

The probability that two balls have the same colour is $\boxed{\frac{29}{44}}$.