Eddie has 16 toy cars, of which 8 are white, 5 are black and 3 are silver. He places all the cars in a bag and selects three of them at random, without replacement.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

1. [1] 1 outcome seen as the product of 3 fractions with 8, 7, 6 or 5, 4, 3 or 3, 2, 1 as numerators and 16, 15, 14 or 16, 16, 16 as denominators Or 1 outcome correct in terms of combinations.
2. [1] Sum of 3 correct identified scenarios (may be identified by correct unsimplified numerator values).
3. [1] 0.1195 < p < 0.120.
4. P(WWW) = (8/16) × (7/15) × (6/14) = 336/3360 = 1/10
5. P(BBB) = (5/16) × (4/15) × (3/14) = 60/3360 = 1/56
6. P(SSS) = (3/16) × (2/15) × (1/14) = 6/3360 = 1/560
7. Total P = 336/3360 + 60/3360 + 6/3360 = 402/3360 = 67/560 ≈ 0.1196
8. [1] One outcome of the form (nCr * mCj * kCk) / TC(Total cars C 3) seen. (e.g. 8C2*5C1 / 16C3)
9. [1] At least one identified outcome fully correct (e.g., 140/560 for 2W1B, or 80/560 for 1W2B, or 120/560 for 1W1B1S).
10. [1] Sum of 3 correct identified scenarios (2W1B, 1W2B, 1W1B1S).
11. [1] 2040/3360, 17/28, 0.607
12. P(2W1B) = (8C2 × 5C1) / 16C3 = (28 × 5) / 560 = 140/560.
13. P(1W2B) = (8C1 × 5C2) / 16C3 = (8 × 10) / 560 = 80/560.
14. P(1W1B1S) = (8C1 × 5C1 × 3C1) / 16C3 = (8 × 5 × 3) / 560 = 120/560.
15. Total P = (140 + 80 + 120) / 560 = 340/560 = 17/28 ≈ 0.607.