Explanation:
(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 : \(^{5}C_{2} \times ^{4}C_{1} + ^{5}C_{2} \times ^{3}C_{1} + ^{4}C_{2} \times ^{5}C_{1} + ^{4}C_{2} \times ^{3}C_{1} + ^{3}C_{2} \times ^{5}C_{1} + ^{3}C_2} \times ^{4}C_{1}\)
= \(40 + 30 + 30 + 18 + 15 + 12 = 145\)
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:
= \(1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)}\)
= \(1 - \frac{6(16)}{7(7^{2} - 1)}\)
= \(1 - \frac{2}{7}\)
= \(0.714\)