A-LevelMathematicsProbability and statistics9709/51 · May/June 2020 · Q58 Marks

On Mondays, Rani cooks her evening meal. She has a pizza, a burger or a curry with probabilities 0.35, 0.44, 0.21 respectively. When she cooks a pizza, Rani has some fruit with probability 0.3. When she cooks a burger, she has some fruit with probability 0.8. When she cooks a curry, she never has any fruit.

(a)Draw a fully labelled tree diagram to represent this information.[2]
(b)Find the probability that Rani has some fruit.[2]
(c)Find the probability that Rani does not have a burger given that she does not have any fruit.[4]

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Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[1] Fully correct labelled tree for meal choices with probabilities: P(Pizza)=0.35, P(Burger)=0.44, P(Curry)=0.21.
[1] Fully correct labelled branches for fruit status with conditional probabilities: P(Fruit|Pizza)=0.3, P(No Fruit|Pizza)=0.7; P(Fruit|Burger)=0.8, P(No Fruit|Burger)=0.2; P(Fruit|Curry)=0, P(No Fruit|Curry)=1.
[1] 0.35×0.3 + 0.44×0.8 (+0)
[1] 0.457
[1] P(not B | not fruit) = P(B'∩F') / P(F')
[1] 0.35×0.7 + 0.21×1 / (1-their(b))
[1] 0.455 / 0.543
[1] 0.838 or 455/543

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