A market-stall fruit crate contains 9 apples, 6 of which are good and 3 of which are bruised. A customer picks out 2 apples at random for a bag, taken in su...
Question 1 Report
A market-stall fruit crate contains 9 apples, 6 of which are good and 3 of which are bruised. A customer picks out 2 apples at random for a bag, taken in succession and not put back, to check the quality before buying.
Use the tree diagram to find the probability that exactly one of the two apples selected is bruised. (4)
Answer Details
"Exactly one bruised" out of two picks can happen in two different orders, good-then-bruised or bruised-then-good, so both branches of the tree diagram leading to exactly one bruised apple must be found and added.
Along the branch where the first apple is good and the second is bruised: \( P(\text{good, bruised}) = \dfrac{6}{9} \times \dfrac{3}{8} = \dfrac{18}{72} \), using 8 remaining apples (3 still bruised) for the second pick [1 mark].
Along the branch where the first apple is bruised and the second is good: \( P(\text{bruised, good}) = \dfrac{3}{9} \times \dfrac{6}{8} = \dfrac{18}{72} \), using 8 remaining apples (6 still good) for the second pick [1 mark].
Since these two orders cannot both happen at once, their probabilities are added: \( \dfrac{18}{72}+\dfrac{18}{72} = \dfrac{36}{72} \) [1 mark], which simplifies to \( \dfrac{1}{2} \) [1 mark].
"Exactly one" is different from "at least one": it excludes the case where both apples are bruised, which is why two separate branches (not all four) are added on the tree diagram.
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