1. **Problem statement:** A bag contains 5 red, 4 blue, and 6 white marbles (total 15). Marbles are drawn without replacement. Find probabilities for various events. 2. **(i) Probability the first marble drawn is not blue:** - Total marbles = 15 - Non-blue marbles = 5 (red) + 6 (white) = 11 - Probability = \frac{11}{15} - Decimal: 0.73 or 73% 3. **(ii) Probability the first red marble drawn is the third marble drawn:** - For the first red marble to be third, first two marbles must be non-red. - Non-red marbles = 4 (blue) + 6 (white) = 10 - Step 1: Probability 1st marble is non-red = \frac{10}{15} - Step 2: Probability 2nd marble is non-red (after 1 non-red removed) = \frac{9}{14} - Step 3: Probability 3rd marble is red (5 red remain) = \frac{5}{13} - Multiply: $$\frac{10}{15} \times \frac{9}{14} \times \frac{5}{13}$$ - Simplify step: $$\frac{\cancel{10}}{\cancel{15}} \times \frac{9}{14} \times \frac{5}{13} = \frac{2}{3} \times \frac{9}{14} \times \frac{5}{13}$$ - Calculate: $$\frac{2}{3} \times \frac{9}{14} = \frac{18}{42} = \frac{3}{7}$$ $$\frac{3}{7} \times \frac{5}{13} = \frac{15}{91} \approx 0.1648$$ - Probability ≈ 16.48% 4. **(iii) Probability the first three marbles drawn are all different colours:** - Total marbles = 15 - Step 1: Probability 1st marble any colour = 1 - Step 2: Probability 2nd marble different colour than 1st: - If 1st is red (5), 2nd must be blue or white (4+6=10), so \frac{10}{14} - Similarly for other colours, but total probability is symmetric. - Step 3: Probability 3rd marble different from first two colours: - After 2 different colours drawn, remaining colour count is 5 (if red left), 4 (blue left), or 6 (white left), depending on first two draws. - Calculate total permutations of 3 different colours: - Number of ways to pick 3 different colours in order = 3! = 6 - Probability: $$\frac{5}{15} \times \frac{4+6}{14} \times \frac{\text{remaining colour count}}{13}$$ - Using exact counts: $$\frac{5}{15} \times \frac{10}{14} \times \frac{4}{13} = \frac{5}{15} \times \frac{10}{14} \times \frac{4}{13} = \frac{200}{2730} = \frac{20}{273} \approx 0.0733$$ - Multiply by 6 permutations: $$6 \times 0.0733 = 0.44$$ - So probability ≈ 44% 5. **(iv) Probability at least one blue marble is drawn among first four marbles:** - Use complement rule: Probability no blue marbles in first 4 draws. - Non-blue marbles = 5 + 6 = 11 - Probability all 4 drawn are non-blue: $$\frac{11}{15} \times \frac{10}{14} \times \frac{9}{13} \times \frac{8}{12}$$ - Simplify step: $$\frac{11}{15} \times \frac{10}{14} \times \frac{9}{13} \times \frac{8}{12}$$ - Calculate numerator: 11 × 10 × 9 × 8 = 7920 - Calculate denominator: 15 × 14 × 13 × 12 = 32760 - Fraction: \frac{7920}{32760} = \frac{33}{136} \approx 0.2426 - Probability at least one blue = 1 - 0.2426 = 0.7574 or 75.74% **Final answers:** - (i) 73% - (ii) 16.48% - (iii) 44% - (iv) 75.74%