Higher November 2020 Unit 1 Q19
19. Dewi has a box containing eleven socks.
Six of the socks are red, four are green and one is yellow.
Early one morning, without switching on the light, Dewi selects two socks at random.
(a) Calculate the probability that the first sock selected is yellow and the second is red. [2]
(b) Calculate the probability that Dewi selects two socks of the same colour. [3]
(c) Calculate the probability that at least one green sock is selected. [3]
| Scheme | Marks | Comments |
|---|---|---|
| 1/11×6/10 or equivalent | M1 | |
| = 6/110 (= 3/55) | A1 | ISW |
| Scheme | Marks | Comments |
|---|---|---|
| 6/11×5/10 + 4/11×3/10 [ + 1/11×0/10] | M2 | FT use of consistent incorrect denominator e.g. 120 Full method for finding P(R, R) + P(G, G) [ + P(Y,Y)] M1 for sight of 6 /11×5/10 or 4/11×3/10 |
| = 42/110 (= 21/55) | A1 | ISW If no marks, SC1 for an answer of 53/121 (method with replacement) |
| Scheme | Marks | Comments |
|---|---|---|
| 1 – 7/11×6/10 or equivalent | M2 | FT use of consistent incorrect denominator e.g. 120 M1 for 7/11×6/10 |
| = 68/110 (= 34/55) | A1 | ISW |
Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| 1 – [6/11×5/10 + 6/11×1/10 + 1/11×6/10] or equivalent | M2 | FT use of consistent incorrect denominator e.g. 120 Full method for finding 1 – [P(R,R) + P(R,Y) + P(Y,R)]. Allow M1 if any one of the three subtracted products is omitted. |
| = 68/110 (= 34/55) | A1 | ISW |
Alternative method 2
| Scheme | Marks | Comments |
|---|---|---|
| 4/11×3/10 + 4/11×7/10 + 7/11×4/10 or equivalent | M2 | FT use of consistent incorrect denominator e.g. 120 Full method for finding P(G,G) + P(G,G’) + P(G’,G). Allow M1 for the sum of any two of these three products NB: P(1st sock green) is equivalent to P(G,G) + P(G,G’) or to P(G,G) + P(G’,G) (i.e. credit cannot be given for only P(1st sock green)=4/11 without considering compound events) |
| = 68/110 (= 34/55) | A1 | ISW |
Alternative method 3
| Scheme | Marks | Comments |
|---|---|---|
| 4/11×3/10 + 4/11×6/10 + 4/11×1/10 + 6/11×4/10 + 1/11×4/10 or equivalent | M2 | FT use of consistent incorrect denominator e.g. 120 Full method for finding P(G,G) + P(G,R) + P(G,Y) + P(R,G) + P(Y,G). Allow M1 for the sum of any two of the following
(i.e. credit cannot be given for only P(1st sock green)=4/11 without considering compound events) |
| = 68/110 (= 34/55) | A1 | ISW |
If no marks,
SC1 for an answer of 72/121 [from 1 – 7/11×7/11]
(method with replacement)