Higher June 2023 Paper 5 Q19
19 A box contains 25 discs.
The discs are either blue or yellow in the ratio 4 : 1.
Two discs are chosen at random from the box without replacement.
Find the probability that the two discs are different colours.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{1}{3}\) oe with correct working | 5 | M1 for blue = 20 and yellow = 5 M3 for \(\frac{5}{25} \times \frac{20}{24} + \frac{20}{25} \times \frac{5}{24}\) oe or M2 for \(\frac{5}{25} \times \frac{20}{24}\) oe or M1 for correct tree diagram or sample space or for \(\frac{5}{25}\) and \(\frac{20}{24}\) or \(\frac{20}{25}\) and \(\frac{5}{24}\) oe seen If 0 scored, SC1 for correct answer with no or insufficient working or for P(B) = \(\frac{4}{5}\) oe and P(Y) = \(\frac{1}{5}\) oe | Correct working” requires evidence of M1 and M3 or convincing alternate approach M1 implied from e.g. \(\frac{20}{25}\) and \(\frac{5}{25}\), [B : Y =] 20: 5 Do not award this mark if they then go on to e.g. use 4 and 1 in working on the tree diagram in both stages for the probabilities but allow the FT method marks for the products or probabilities For M3, M2 allow evaluated products e.g. for M3 allow \(\frac{100}{600} + \frac{100}{600}\) oe provided tree diagram given with individual probabilities shown. M2 not awarded if part of a larger product of probabilities For M3, M2, M1 FT their blue and yellow e.g. blue 4 and yellow 1 M3 for \(\frac{4}{5} \times \frac{1}{4} + \frac{1}{5} \times \frac{4}{4}\) [answer \(\frac{2}{5}\) oe] M2 for \(\frac{4}{5} \times \frac{1}{4}\) or \(\frac{1}{5} \times \frac{4}{4}\) M1 for \(\frac{4}{5}\) and \(\frac{1}{4}\) |