Freddie has two bags of marbles. Bag X contains 7 red marbles and 3 blue marbles. Bag Y contains 4 red marbles and 1 blue marble. Freddie chooses one of the bags at random. A marble is removed at random from that bag and not replaced. A new red marble is now added to each bag. A second marble is then removed at random from the same bag that the first marble had been removed from.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [B1] 1st column, 2 branches identified X, Y with probabilities ½, ½ indicated.
2. [B1] 2nd column (1st marble pick) of 4 branches identified R B R B (oe) and probabilities 7/10, 3/10, 4/5, 1/5 indicated appropriately.
3. [B1] 3rd column (2nd marble pick) of 8 branches identified R B R B R B R [B] (oe) and probabilities 7/10, 3/10, 8/10, 2/10, 4/5, 1/5, 5/5, 0 indicated appropriately. Condone omission of YBB branch if YBR branch is fully correct. Ignore any additional columns of branches. If separate tree diagrams for bags X and Y, B0B1B1 max if bags clearly identified.
4. [B1 FT] [P(BB)=] (1/2)×(3/10)×(2/10) + (1/2)×(1/5)×0 = 6/200 seen. Accept unsimplified. FT from 6(a) unsimplified only with 3 term probabilities.
5. [B1 FT] Either [P(XRR) =] (1/2)×(7/10)×(7/10) or [P(YRR) =] (1/2)×(4/5)×(4/5) seen. FT from 6(a) unsimplified only with 3 term probabilities.
6. [M1] [P(BB) + P(XRR) + P(YRR) =] their 6/200 + their 49/200 + their 16/50. Accept unsimplified, consistent with tree diagram if not clearly identified by notation.
7. [A1] 119/200, 0.595.
8. [M1] FT from their 6(a) and their 6(b) with 3 term probabilities unsimplified only or correct. Accept (1/2)×(1/5)×(1/5) + (1/2)×(4/5)×(1/5) / (1 - their 6(b)). Accept (4/81) / (36/81) or (0.08+0.1) / (0.405).
9. [A1] 9/50, 4/81, 0.444. Accept 36/81 or 0.4.