Higher November 2018 Paper 2 Q18
18 A bag contains 20 discs.
10 are red, 7 are blue and 3 are green.
(a) Marnie takes a disc at random before putting it back in the bag.
Nick then takes a disc at random before putting it back in the bag.
Olly then takes a disc at random.
Work out the probability that they all take a red disc. [2 marks]
(b) All 20 discs are in the bag.
Reggie takes three discs at random, one after the other.
After he takes a disc he does not put it back in the bag.
Reggie’s first disc is blue.
Work out the probability that all three discs are different colours. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{10}{10 + 7 + 3}\) or \(\dfrac{10}{20}\) or \(\dfrac{5}{10}\) or \(\dfrac{1}{2}\) or 0.5 | M1 | oe eg 50% |
| \(\dfrac{1}{8}\) or 0.125 or 12.5% | A1 | oe eg \(\dfrac{1000}{8000}\) or \(\dfrac{125}{1000}\) |
Additional guidance
| Ignore incorrect conversion if correct answer seen | |
| Answer \(\dfrac{1}{2}\) | M1 |
| 10 out of 20 | M0 |
| 10 : 20 | M0 |
| Answer 1 out of 8 | M1A0 |
| Answer 1 : 8 is A0 but M1 is possible | |
| \(\dfrac{10}{20} \quad \dfrac{7}{20} \quad \dfrac{3}{20}\) | M1 |
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{10}{19}\) or \(\dfrac{3}{19}\) | M1 | oe allow [0.52, 0.53] or [0.15, 0.16] |
| \(\dfrac{10}{19} \times \dfrac{3}{18}\) \((\times 2)\) or \(\dfrac{3}{19} \times \dfrac{10}{18}\) \((\times 2)\) or \(\dfrac{5}{57}\) \((\times 2)\) or [0.087, 0.088] \((\times 2)\) | M1dep | oe eg \(1 \times \dfrac{10}{19} \times \dfrac{3}{18}\) or \(\dfrac{30}{342}\) allow \([0.52, 0.53] \times [0.16, 0.17]\) or \([0.15, 0.16] \times [0.55, 0.56]\) |
| \(\dfrac{10}{57}\) or 0.175… or 17.5…% | A1 | oe eg \(\dfrac{60}{342}\) SC2 \(\dfrac{7}{38}\) or 0.184… oe |
Additional guidance
| \(\dfrac{7}{20} \times \dfrac{10}{19} \times \dfrac{3}{18}\) | M1M0A0 |
| \(\dfrac{7}{20} \times \dfrac{3}{19} \times \dfrac{10}{18}\) | M1M0A0 |
| If more than one product is seen, the correct one(s) must be selected for 2nd M1 \(\dfrac{10}{19} \times \dfrac{6}{18} + \dfrac{3}{19} \times \dfrac{10}{18}\) | M1M0A0 |
| Both correct products selected but multiplied together scores M1 only \(\dfrac{10}{19} \times \dfrac{3}{18} \times \dfrac{3}{19} \times \dfrac{10}{18}\) | M1M0A0 |
| Ignore incorrect conversion if correct answer seen | |
| 5 out of 57 cannot score 2nd M1 but implies 1st M1 | |
| 5 : 57 cannot score 2nd M1 but 1st M1 is possible | |
| Answer 10 out of 57 | M1M1A0 |
| Answer 10 : 57 is A0 but M2 or M1M0 is possible |