Higher June 2023 Paper 2 Q20
20 There are 12 counters in a bag.
3 of the counters are red
9 of the counters are green
Ameya, Jack and Ella each take at random one counter from the bag.
Work out the probability that at least one red counter is still in the bag.
(3)
| Scheme | Marks |
|---|---|
(RRR:) \(\dfrac{3}{12} \times \dfrac{2}{11} \times \dfrac{1}{10}\;\left(= \dfrac{1}{220}\right)\) or (0.25 × 0.18… × 0.1 = 0.0045…) (2R, 1G:) \(\dfrac{3}{12} \times \dfrac{2}{11} \times \dfrac{9}{10}\;\left(= \dfrac{9}{220}\right)\) or (0.25 × 0.18… × 0.9 = 0.0409…) (2G, 1R:) \(\dfrac{3}{12} \times \dfrac{9}{11} \times \dfrac{8}{10}\;\left(= \dfrac{36}{220} = \dfrac{9}{55}\right)\) or (0.25 × 0.81… × 0.8 = 0.163…) (GGG:) \(\dfrac{9}{12} \times \dfrac{8}{11} \times \dfrac{7}{10}\;\left(= \dfrac{84}{220} = \dfrac{21}{55}\right)\) (0.25 × 0.72… × 0.7 = 0.381…) | M1oe |
\(1 - \text{``}{\dfrac{1}{220}}\text{''}\) (1 – “0.0045….”) or \(\text{``}{\dfrac{84}{220}}\text{''} + 3 \times \text{``}{\dfrac{36}{220}}\text{''} + 3 \times \text{``}{\dfrac{9}{220}}\text{''}\) (0.381… + 3 × 0.163… + 3 × 0.0409…) | M1oe |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{219}{220}\) | A1oe |
| (3) | |
| (3 marks) |
Notes
M1oe: For an expression to find one of the stated probabilities
M1oe: Dep M1 Complete method
A1oe: 0.9954….allow 0.99 (99%) or 0.995 (99.5%)