(a) A bag contains 5 blue, 4 green and 3 yellow balls. All the balls

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)

Spearman's rank correlation cofficient:

= \(1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)}\)

= \(1 - \frac{6(16)}{7(7^{2} - 1)}\)

= \(1 - \frac{2}{7}\)

= \(0.714\)