Chen has three boxes. Box A contains 5 counters, of which 3 are white and 2 are yellow. Box W contains 4 red marbles and 3 blue marbles. Box Y contains 5 red marbles and 3 blue marbles. Chen chooses one counter at random from box A. If the counter is white, he chooses two marbles from box W, at random and without replacement. If the counter is yellow, he chooses two marbles from box Y, at random and without replacement.
- (a)Draw a fully labelled tree diagram to illustrate this information, including all the probabilities.[3]
- (b)Find the probability that one red marble and one blue marble are obtained.[3]
- (c)Find the probability that a white counter is chosen from box A, given that one red marble and one blue marble are obtained.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.
- [B1] Correct tree diagram structure. (3 tiers).
- [B1] Counter and first marble probabilities correct with outcomes labelled.
- [B1] Second marble probabilities correct with outcomes labelled.
- [M1] At least 2 correct 3-term products linked to correct scenarios. May be clearly identified on tree diagram. FT from a 3-tier tree.
- [M1] 4 correct identified scenarios added. May be clearly identified on the tree diagram, condone identification by correct (or FT) un-simplified products.
- [A1] SCB1 for 39/70 if one or both M marks not scored WWW.
- [M1] Correct formula for conditional probability using their probabilities from 3-tier tree with their (b) < 1 or correct in denominator. Condone 12/35 or their P(WRB)+P(WBR) from part b / their 39/70 or unsimplified expression.
- [A1] 8/13 = 0.615
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