(a) 5b, 4g, 3y = 12 balls.
(i) p(all 3 balls have the same colour) = p(all 3 blue or all 3 green or all 3 yellow)
= $\frac{5}{12}\times \frac{4}{11}\times \frac{3}{10}+\frac{4}{12}\times \frac{3}{11}\times \frac{2}{10}+\frac{3}{12}\times \frac{2}{11}\times \frac{1}{10}$
= $\frac{60}{1320}+\frac{24}{1320}+\frac{6}{1320}$
= $\frac{90}{1320}=\frac{3}{44}$
(ii) Two balls of the same colour :
Sample space for selecting with restriction (2 blue 1 green or 2 blue 1 yellow or 2 green 1 blue or 2 green 1 yellow or 2 yellow 1 blue or 2 yellow 1 green)
Ways : $\text{Extra close brace or missing open brace}$
= $40+30+30+18+15+12=145$
Selecting without restriction 3 balls out of 12 balls : ${}^{12}{C}_3$
= $\frac{12!}{3!9!}$
= $220$
Probability = $\frac{\text{selection with restriction}}{\text{selection without restriction}}=\frac{145}{220}$
= $\frac{29}{44}$
(b)

$R_P$	$R_C$	$d=R_P-R_C$	$d^2$
6	7	-1	1
5	6	-1	1
4	2	2	4
3	4	-1	1
2	1	1	1
7	5	2	4
1	3	-2	4
			16

Spearman's rank correlation cofficient:
= $1-\frac{6\sum d^2}{n(n^2-1)}$
= $1-\frac{6(16)}{7(7^2-1)}$
= $1-\frac{2}{7}$
= $0.714$