Higher November 2018 Unit 1 Q17
17. A box contains 4 red balls, 5 yellow balls and 1 green ball.
Two balls are to be chosen at random, without replacement.
(a) Find the probability of choosing 1 red ball and 1 green ball. [3]
(b) Find the probability that the two balls chosen will not be the same colour. [4]
| Scheme | Marks | Comments |
|---|---|---|
| sight of 4/10 × 1/9 OR 1/10 × 4/9 | M1 | |
| 4/10 × 1/9 + 1/10 × 4/9 OR 4/10 × 1/9 × 2 | M1 | Complete correct method. |
| = 8/90 (= 4/45) | A1 | ISW SC1 for method with replacement, leading to an answer of 8/100. |
| Scheme | Marks | Comments |
|---|---|---|
| 1 – P(both the same colour) | M1 | FT a repeated incorrect total from part (a). |
| = 1 – [5/10 × 4/9 + 4/10 × 3/9] (= 1 – 32/90) | M2 | Complete correct method. M1 for 1 numerical slip. Accept presence of a correct product of 1/10 × 0/9 for P(GG). |
| = 58/90 (= 29/45) | A1 | ISW. If no marks awarded, SC1 for sight of 32/90 (probability of both the same colour) SC1 for an answer of 70/90 (from 1 – P(Y,Y)) SC1 for an answer of 78/90 (from 1 – P(R,R)) SC2 for method with replacement, leading to an answer of 58/100. SC1 for method with replacement, without a related answer or with 1 numerical slip or without considering P(GG) (59/100). |
Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
| P(RY or YR or YG or GY or GR or RG) | M1 | |
| = 4/10×5/9 + 5/10×4/9 + 5/10×1/9 + 1/10×5/9 + 1/10×4/9 + 4/10×1/9 or equivalent | M2 | Complete correct method. M1 for 1 numerical slip. FT ‘their part (a)’ for P(RG) + P(GR). |
| = 58/90 (= 29/45) | A1 | ISW If no marks awarded, SC2 for this method and related answer, having omitted one product (out of 6) SC2 for an answer of 29/90 (from P(RY or YG or GR)) SC1 for this method, having omitted one product, with no related correct answer SC1 for this method and related answer, having omitted two products |
Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| P(RR’ or YY’ or GG’) | M1 | |
| = 4/10×6/9 + 5/10×5/9 + 1/10(×9/9) | M2 | Complete correct method. M1 for 1 numerical slip or M1 if ‘doubling’ is seen (at any stage) |
| = 58/90 (= 29/45) | A1 | ISW If no marks awarded, SC1 for this method and related answer, having omitted one product (out of 3) |
Corrected by Westie's Workshop
- The board's mark scheme printed “70/90 (from 1 – P(R,R)) and 78/90 (from 1 – P(Y,Y))”. Changed to “70/90 (from 1 – P(Y,Y)) and 78/90 (from 1 – P(R,R))” because with 4 red and 5 yellow balls, P(Y,Y) = 5/10 × 4/9 = 20/90, so 1 – P(Y,Y) = 70/90, and P(R,R) = 4/10 × 3/9 = 12/90, so 1 – P(R,R) = 78/90. The labels were the wrong way round.
Original: see the board's mark scheme PDF (link above).