Higher June 2024 Paper 2 Q18
18 There are only 4 red counters, 3 yellow counters and 1 green counter in a bag.
Tony takes at random three counters from the bag.
Work out the probability that there are now more yellow counters than red counters in the bag.
You must show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{5}{28}\) | P1 | for a correct probability for 2nd or 3rd counter, eg \(\dfrac{3}{7}\) or \(\dfrac{4}{7}\) or \(\dfrac{1}{7}\) or \(\dfrac{1}{6}\) or \(\dfrac{2}{6}\) or \(\dfrac{3}{6}\) |
| P1 | for a correct product for 2 red and 1 green eg, \((\text{P(RRG)} =)\ \dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{1}{6}\ \left(= \dfrac{12}{336} \text{ or } \dfrac{1}{28}\right)\) or \((\text{P(RGR)} =)\ \dfrac{4}{8} \times \dfrac{1}{7} \times \dfrac{3}{6}\ \left(= \dfrac{12}{336} \text{ or } \dfrac{1}{28}\right)\) or \((\text{P(GRR)} =)\ \dfrac{1}{8} \times \dfrac{4}{7} \times \dfrac{3}{6}\ \left(= \dfrac{12}{336} \text{ or } \dfrac{1}{28}\right)\) | |
| P1 | for a correct product for 3 red, eg \((\text{P(RRR)} =)\ \dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{2}{6}\ \left(= \dfrac{24}{336} \text{ or } \dfrac{2}{28}\right)\) | |
| P1 | for a complete process, eg \(\left(\dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{1}{6}\right) + \left(\dfrac{4}{8} \times \dfrac{1}{7} \times \dfrac{3}{6}\right) + \left(\dfrac{1}{8} \times \dfrac{4}{7} \times \dfrac{3}{6}\right) + \left(\dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{2}{6}\right)\) | |
| A1 | for \(\dfrac{60}{336}\) oe eg \(\dfrac{5}{28}\) SCB2 if P0 scored for answer of \(\dfrac{112}{512}\) oe (replacement) |
Additional guidance
Accept equivalent fractions, decimals (0.17... or 0.18) or percentages (17% or 18%)