A bag contains 5 blue balls, 4 green balls, and 3 yellow balls. Three balls are drawn at random without replacement. What is the probability that all three balls are the same colour?

Total number of balls = $5 \text{ (blue)} + 4 \text{ (green)} + 3 \text{ (yellow)} = 12$ balls.
The total number of ways to draw 3 balls from 12 without replacement is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

• 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: $\binom{3}{3} = 1$ Total ways for all three balls to have the same colour = $10 + 4 + 1 = 15$.

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

a) ii. Two balls have the same colour.

• 2 Blue and 1 Green: $\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40$
• 2 Blue and 1 Yellow: $\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30$
• 2 Green and 1 Blue: $\binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30$
• 2 Green and 1 Yellow: $\binom{4}{2} \times \binom{3}{1} = 6 \times 3 = 18$
• 2 Yellow and 1 Blue: $\binom{3}{2} \times \binom{5}{1} = 3 \times 5 = 15$
• 2 Yellow and 1 Green: $\binom{3}{2} \times \binom{4}{1} = 3 \times 4 = 12$ Total ways for two balls to have the same colour = $40 + 30 + 30 + 18 + 15 + 12 = 145$.

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

b) Calculate the Spearman's rank correlation coefficient.

The number of candidates, $n = 7$.