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Question 105 of 108 in Unit 10

Unit 10

10 marks · Hard

A fruit seller is combining baskets of apples and pears. The amount of apples and pears after a trade can be represented as algebraic fractions, but the trade involves uncertain conditions shown by a probability tree.

(a)

Express the following as a single fraction in its simplest form:

(b)

A customer selects a basket at random. There are 3 baskets labelled A, B and C. The probability that she selects basket A is 0.5. If she does not select A, the probability she selects B is 0.7.

i) Draw a tree diagram to show all possible outcomes.

[4 marks]
(c)

ii) Find the probability that she selects basket C.

[6 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) ((2x+1)(x-3) + (x-1)(x+2))/((x+2)(x-3)) = (3x^2 - 3x - 1)/((x+2)(x-3)) · must show Correct calculation of numerator, Fully simplified single fraction

(b) Tree diagram showing branches for selecting A, then B, then C, with all probabilities labelled · must show First split: A(0.5), Not A(0.5), Second split after Not A: B(0.7), C(0.3), All branch probabilities shown

(c) 0.15 · must show Probability of not A is 0.5, probability of C given not A is 0.3, 0.5 * 0.3 = 0.15

Common mistakes: Expanding or simplifying numerator incorrectly; Incorrect lowest common denominator; Wrong probabilities or missing branches in tree diagram; Incorrect sum for probability of C

Show mark scheme and worked solution

Marking points

M1 (a(b)) · 2 marks · Finds common denominator and rewrites numerators correctly

A1 (a(b)) · 2 marks · Expands brackets and simplifies numerator fully (follow-through allowed) (follow-through)

M2 (a(c)) · 1 mark · Initial split: probability for A and not A

M3 (a(c)) · 2 marks · Branches for B and C with correct probabilities

M4 (a(c)) · 2 marks · Identifies correct branches for selecting C (follow-through allowed) (follow-through)

A2 (a(c)) · 1 mark · Correct calculation 0.5 x 0.3 = 0.15 (follow-through allowed) (follow-through)

Worked solution

a) with common denominator :

So final answer:

b) Tree: Start with A (0.5), Not A (0.5). Not A splits into B (0.7), C (0.3).

c) P(C) = 0.5 (probability not A) × 0.3 (probability C after not A) = 0.15.

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