A factory produces chocolates in three flavours: lemon, orange and strawberry in the ratio 3 : 5 : 7 respectively. Nell checks the chocolates on the production line by choosing chocolates randomly one at a time.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

1. B1 0.0524288 rounded to more than 3SF if final answer
2. P(Lemon) = 3/15 = 1/5, P(Not Lemon) = 12/15 = 4/5
3. P(7th is first lemon) = (4/5)⁶ × (1/5) = 0.0524288
4. M1 (1-4/5⁶). FT their 1/5 or correct. From final answer
5. A1 0.262144 rounded to more than 3SF
6. P(first lemon is after at least 6 chocolates) = P(first 6 are not lemon) = (4/5)⁶ = 0.262144
7. M1 (10/15) × (9/14) × (8/13) , no additional terms OR M1 10C3 / 15C3
8. A1 0.263736 rounded to more than 3SF
9. P(none orange) = (10/15) × (9/14) × (8/13) = 24/91 ≈ 0.263736
10. M1 All probabilities of the form: 7/a × 5/b × 3/c, 13 ≤ a,b,c ≤ 15 OR M1 ³C₁ × ⁵C₁ × ⁷C₁ / ¹⁵C₃
11. M1 e/f × g/h × i/j × 3! e,f,g,h,i,j positive integers forming probabilities or 6 identical probability calculations or values added, no additional terms
12. A1 0.230769 rounded (not truncated) to more than 3SF
13. P(different flavours) = (³C₁ × ⁵C₁ × ⁷C₁) / ¹⁵C₃ = 105 / 455 = 3/13 ≈ 0.230769
14. B1 ⁷C₂׳C₁ seen (numerator for 2S, 1L) OR B1 7/15 × 6/14 × 3/13 × 3 seen (for 2S1L using sequential)
15. B1 ⁷C₃ seen in numerator (for 3S) OR B1 7/15 × 6/14 × 5/13 seen in numerator (for 3S using sequential)
16. M1 Fraction with ¹⁰C₃ or consistent with their numerator of 6(c) in denominator
17. A1 Accept 0.816
18. P(none orange) = 10C3 / 15C3 = 120/455 = 24/91
19. P(at least 2S and no orange):
20. 2S1L: ⁷C₂ × ³C₁ = 21 × 3 = 63
21. 3S: ⁷C₃ = 35
22. Total ways for numerator = 63 + 35 = 98
23. P(at least 2S | none orange) = (98/455) / (120/455) = 98/120 = 49/60 ≈ 0.81666...