Higher June 2025 Paper 2 Q4
4. Yasmin has two bags, Bag \(A\) and Bag \(B\).
Each bag contains some red counters and some blue counters.
Bag \(A\) contains 2 red counters and 3 blue counters.
In Bag \(B\) the number of red counters is three times the number of blue counters.
Yasmin randomly chooses one counter from Bag \(A\) and then one counter from Bag \(B\).
(a) Complete the probability tree diagram. [3]
(b) Calculate the probability that Yasmin chooses at least one blue counter. [3]
| Scheme | Marks | Comments |
|---|---|---|
| Bag A branches completed correctly \(\frac{2}{5}\) and \(\frac{3}{5}\) oe or 0.4 and 0.6 | B1 | Allow the use of percentages |
| Both pairs of Bag B branches completed correctly \(\frac{3}{4}\) and \(\frac{1}{4}\) AND \(\frac{3}{4}\) and \(\frac{1}{4}\) oe or 0.75 and 0.25 AND 0.75 and 0.25 | B2 | B1 for one of the following:
|
| Scheme | Marks | Comments |
|---|---|---|
| (P(RB) + P(BR) + P(BB) =) \(\frac{2}{5}\times\frac{1}{4} + \frac{3}{5}\times\frac{3}{4} + \frac{3}{5}\times\frac{1}{4}\) oe (\(\frac{2}{20} + \frac{9}{20} + \frac{3}{20}\) or \(0.1 + 0.45 + 0.15\)) or (1 – P(RR) =) \(1 - \frac{2}{5}\times\frac{3}{4}\) oe (\(1 - \frac{6}{20}\) or \(1 - 0.3\)) or (P(RB) + P(B) =) \(\frac{2}{5}\times\frac{1}{4} + \frac{3}{5}\) oe (\(\frac{2}{20} + \frac{3}{5}\) or \(0.1 + 0.6\)) | M2 | Allow a restart in (b) FT from (a) provided at least B1 awarded and the values on every pair of branches add to 1. M1 for any one of the following:
|
| \(= \frac{14}{20}\) oe or 0.7 ISW | A1 | FT from M2 only provided a numeric value given |
| (6) |
Corrected by Westie's Workshop
- The board's mark scheme printed “(= 6/20 or 0.3)1”. Changed to “(= 6/20 or 0.3)” because a stray “1” was printed after the bracket; the value is 0.3.
Original: see the board's mark scheme PDF (link above).