(a) A bag contains 5 blue, 4 green and 3 yellow balls. All the balls are identical except for colour. Three balls are drawn at random without replacement. Find the probability that : (i) all three balls have the same colour ; (ii) two balls have the same colour.
(b) The table shows the ranks of the marks scored by 7 candidates in Physics and Chemistry tests.
| Physics | 6 | 5 | 4 | 3 | 2 | 7 | 1 |
| Chemistry | 7 | 6 | 2 | 4 | 1 | 5 | 3 |
Calculate the Spearman’s rank correlation coefficient.
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:
= \(1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)}\)
= \(1 - \frac{6(16)}{7(7^{2} - 1)}\)
= \(1 - \frac{2}{7}\)
= \(0.714\)