This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
a) i. All three balls have the same colour. Step 1: Calculate the total number of balls and the total number of ways to draw 3 balls. Total number of balls = 5 (blue) + 4 (green) + 3 (yellow) = 12 balls. The total number of ways to draw 3 balls from 12 without replacement is given by the combination formula nk = (n!)/(k!(n-k)!). Total ways = 123 = (12 × 11 × 10)/(3 × 2 × 1) = 2 × 11 × 10 = 220 Step 2: Calculate the number of ways to draw three balls of the same colour. • Number of ways to draw 3 blue balls: 53 = (5 × 4 × 3)/(3 × 2 × 1) = 10 • Number of ways to draw 3 green balls: 43 = (4 × 3 × 2)/(3 × 2 × 1) = 4 • Number of ways to draw 3 yellow balls: 33 = 1 Total ways for all three balls to have the same colour = 10 + 4 + 1 = 15. Step 3: Calculate the probability. P(all three same colour) = Number of ways for same colourTotal ways = (15)/(220) = (3)/(44) The probability that all three balls have the same colour is (3)/(44). a) ii. Two balls have the same colour. Step 1: Calculate the number of ways to draw exactly two balls of one colour and one ball of a different colour. • 2 Blue and 1 Green: 52 × 41 = 10 × 4 = 40 • 2 Blue and 1 Yellow: 52 × 31 = 10 × 3 = 30 • 2 Green and 1 Blue: 42 × 51 = 6 × 5 = 30 • 2 Green and 1 Yellow: 42 × 31 = 6 × 3 = 18 • 2 Yellow and 1 Blue: 32 × 51 = 3 × 5 = 15 • 2 Yellow and 1 Green: 32 × 41 = 3 × 4 = 12 Total ways for two balls to have the same colour = 40 + 30 + 30 + 18 + 15 + 12 = 145. Step 2: Calculate the probability. P(two balls same colour) = Number of ways for two same colourTotal ways = (145)/(220) = (29)/(44) The probability that two balls have the same colour is (29)/(44). b) Calculate the Spearman's rank correlation coefficient. Step 1: List the ranks and calculate the differences (d_i) and squared differences (d_i^2). The number of candidates, n = 7. | Candidate | Physics Rank (R_1) | Chemistry Rank (R_2) | d_i = R_1 - R_2 | d_i^2 | | :-------- | :------------------- | :--------------------- | :---------------- | :------ | | 1 | 6 | 7 | -1 | 1 | | 2 | 5 | 6 | -1 | 1 | | 3 | 4 | 2 | 2