(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)