Higher January 2020 Paper 1 Q18
18 There are 16 sweets in a bowl.
4 of the sweets are blackcurrant.
5 of the sweets are lemon.
7 of the sweets are orange.
Anna, Ravi and Sam each take at random one sweet from the bowl.
Work out the probability that the 5 lemon sweets are still in the bowl.
(4)
| Scheme | Marks |
|---|---|
\(\dfrac{4}{16} \times \dfrac{3}{15} \times \dfrac{2}{14} \left(= \dfrac{24}{3360} = \dfrac{1}{140}\right)\) oe or \(\dfrac{7}{16} \times \dfrac{6}{15} \times \dfrac{5}{14} \left(= \dfrac{210}{3360} = \dfrac{1}{16}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{4}{15} \times \dfrac{3}{14} \left(= \dfrac{60}{3360} = \dfrac{1}{56}\right)\) oe | M1 |
\(\dfrac{4}{16} \times \dfrac{7}{15} \times \dfrac{6}{14} \left(= \dfrac{168}{3360} = \dfrac{1}{20}\right)\) oe or \(\dfrac{4}{16} \times \dfrac{3}{15} \times \dfrac{7}{14} \left(= \dfrac{84}{3360} = \dfrac{1}{40}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{4}{15} \times \dfrac{4}{14} \left(= \dfrac{80}{3360} = \dfrac{1}{42}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{4}{15} \times \dfrac{7}{14} \left(= \dfrac{140}{3360} = \dfrac{1}{24}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{4}{15} \times \dfrac{3}{14} \left(= \dfrac{60}{3360} = \dfrac{1}{56}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{7}{15} \times \dfrac{6}{14} \left(= \dfrac{210}{3360} = \dfrac{1}{16}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{7}{15} \times \dfrac{4}{14} \left(= \dfrac{140}{3360} = \dfrac{1}{24}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{4}{15} \times \dfrac{11}{14} \left(= \dfrac{220}{3360} = \dfrac{11}{168}\right)\) oe or \(\dfrac{5}{16} \times \dfrac{11}{15} \times \dfrac{10}{14} \left(= \dfrac{550}{3360} = \dfrac{55}{336}\right)\) oe | M1 |
\(\text{‘}\dfrac{24}{3360}\text{’} + 3 \times \text{‘}\dfrac{84}{3360}\text{’} + \text{‘}\dfrac{210}{3360}\text{’} + 3 \times \text{‘}\dfrac{168}{3360}\text{’}\) oe or \(1 - \left(\text{‘}\dfrac{60}{3360}\text{’} + 3 \times \text{‘}\dfrac{80}{3360}\text{’} + 3 \times \text{‘}\dfrac{140}{3360}\text{’} + 3 \times \text{‘}\dfrac{60}{3360}\text{’} + 3 \times \text{‘}\dfrac{210}{3360}\text{’} + 6 \times \text{‘}\dfrac{140}{3360}\text{’}\right)\) oe or \(1 - \left(\text{‘}\dfrac{60}{3360}\text{’} + 3 \times \text{‘}\dfrac{220}{3360}\text{’} + 3 \times \text{‘}\dfrac{550}{3360}\text{’}\right)\) oe | M1 |
| \(\dfrac{990}{3360}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for finding \(BBB\) or \(OOO\) or \(LLL\)
OR M3 (for the three M1 marks) for \(\dfrac{11}{16} \times \dfrac{10}{15} \times \dfrac{9}{14}\) oe
M1: for finding the following in any order
\(BOO\) or \(BBO\)
or
\(LLB\) or \(LLO\) or \(LBB\) or \(LOO\) or \(LOB\)
or
\(LLX\) or \(LXX\) (\(X\) = not \(L\))
M1: for a complete method
A1: for \(\dfrac{990}{3360}\) oe e.g. \(\dfrac{33}{112}\) or 0.29(464...)
(corrected from the printed mark scheme: the second complete-method expression is printed without the + signs before “3 × ‘210/3360’” and “6 × ‘140/3360’”)