Higher June 2025 Paper 1 Q21
21 A box contains 20 counters.
9 of the counters are red
7 of the counters are yellow
4 of the counters are green
Alex takes at random three counters from the box.
Work out the probability that exactly two of the three counters are the same colour.
(3)
| Scheme | Marks |
|---|---|
eg (P(RRY) =) \(\dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18}\left(= \dfrac{504}{6840} = \dfrac{7}{95}\right)\) oe or (P(RRG) =) \(\dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{4}{18}\left(= \dfrac{288}{6840} = \dfrac{4}{95}\right)\) oe or (P(RRR) =) \(\dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18}\left(= \dfrac{504}{6840} = \dfrac{7}{95}\right)\) oe or (P(RRR′) =) \(\dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{11}{18}\left(= \dfrac{792}{6840} = \dfrac{11}{95}\right)\) oe or (P(YYR) =) \(\dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{9}{18}\left(= \dfrac{378}{6840} = \dfrac{21}{380}\right)\) oe or (P(YYG) =) \(\dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{4}{18}\left(= \dfrac{168}{6840} = \dfrac{7}{285}\right)\) oe or (P(YYY) =) \(\dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{5}{18}\left(= \dfrac{210}{6840} = \dfrac{7}{228}\right)\) oe or (P(YYY′) =) \(\dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{13}{18}\left(= \dfrac{546}{6840} = \dfrac{91}{1140}\right)\) oe or (P(GGR) =) \(\dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{9}{18}\left(= \dfrac{108}{6840} = \dfrac{3}{190}\right)\) oe or (P(GGY) =) \(\dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{7}{18}\left(= \dfrac{84}{6840} = \dfrac{7}{570}\right)\) oe or (P(GGG) =) \(\dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{2}{18}\left(= \dfrac{24}{6840} = \dfrac{1}{285}\right)\) oe or (P(GGG′) =) \(\dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{16}{18}\left(= \dfrac{192}{6840} = \dfrac{8}{285}\right)\) oe or (P(RGY) =) \(\dfrac{9}{20} \times \dfrac{7}{19} \times \dfrac{4}{18}\left(= \dfrac{252}{6840} = \dfrac{7}{190}\right)\) oe | M1 |
(P(RRR′ or YYY′ or GGG′) =) \(\left(3 \times \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{11}{18}\right) + \left(3 \times \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{13}{18}\right) + \left(3 \times \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{16}{18}\right)\) oe or (P(RRY or RRG or YYR or YYG or GGR or GGY) =) \(\left(3 \times \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18}\right) + \left(3 \times \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{4}{18}\right) + \left(3 \times \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{9}{18}\right) + {}\) \(\left(3 \times \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{4}{18}\right) + \left(3 \times \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{9}{18}\right) + \left(3 \times \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{7}{18}\right)\) oe or (1 – P(RRR or YYY or GGG or RGY) =) \(1 - \left(\left(\dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18}\right) + \left(\dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{5}{18}\right) + \left(\dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{2}{18}\right) + \left(6 \times \dfrac{9}{20} \times \dfrac{7}{19} \times \dfrac{4}{18}\right)\right)\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) SCB1 for an answer of \(\dfrac{669}{1000}\) oe eg 0.66(9) or 66(.9)% Answer: \(\dfrac{51}{76}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for finding one correct product, does not need to be labelled
or
for an answer of \(\dfrac{17}{76}\) oe eg 0.22(3…) or 22(.3…)%
or \(\dfrac{65}{76}\) oe eg 0.85(5…) or 85(.5…)%
M1: for a complete calculation
A1: oe eg 0.67(1…) or 67(.1…)%