Bag A contains 6 red marbles, 5 blue marbles and 1 green marble. Bag B contains 5 red marbles and 3 blue marbles. A marble is chosen at random from bag A and placed in bag B. A marble is now chosen at random from bag B.

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [B1] Correct structure and probabilities for Bag A branches which must be the first set of branches, including labels R, B, G. (Probabilities for Bag A selection R,B,G (6/12, 5/12, 1/12))
2. [B1] Correct structure and probabilities for one Bag B branch, including labels R, B, [G]. Condone inclusion of branches with 0 probability.
3. [B1] Correct structure and probabilities for remaining Bag B branches, including labels R, B, [G]. Condone inclusion of branches with 0 probability. Additional branches without a probability of zero lose this mark.
4. [M1] [P(RR) + P(BB) + P(GG) =] (6/12 * 6/9) + (5/12 * 4/9) + (1/12 * 1/9). Correct or FT tree diagram probabilities with all 3 (R, B and G) branches. Note: working may be by tree diagram.
5. [A1] = 57/108 or 19/36. Accept 0.5278, 0.528.
6. [M1] P(A blue | B blue) = P(A blue and B blue) / P(B blue) (5/12 * 4/9) or (5/12 * 20/27) or their P(BB) from Q5(b) seen as a numerator of a single fraction. FT tree diagram with values from only 2 bags.
7. [B1 FT] Denominator P(B blue) = (6/12 * 3/9) + (5/12 * 4/9) + (1/12 * 3/9). (6/12 * 3/9) + (5/12 * 4/9) + (1/12 * 3/9) seen as the denominator of a single fraction. (5/27 or their P(BB) may be seen from Q5b in place of 5/12 * 4/9). FT tree diagram with values from only 2 bags.
8. [A1] = 20/41. Accept 0.488, 0.4878. (SCB1 if either M or B mark not scored 20/41 or 0.4878.)