Unit 8

8 marks · Normal

This question involves composite functions and probability tree diagrams.

(a)

Let the function be defined by and the function by .

(i)

Write down an expression for the composite function and simplify your answer.

[2 marks]

(ii)

Calculate the value of .

[1 mark]

(b)

A bag contains 5 red and 3 blue balls. A ball is chosen at random, noted, and replaced. A second ball is chosen. Tree diagrams are to be used for this part.

(i)

Draw a tree diagram showing all possibile outcomes and label the probabilities on each branch.

[2 marks]

(ii)

Work out the probability that both balls chosen are red.

[1 mark]

(iii)

Work out the probability that one ball is red and one is blue (in any order).

[2 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(i)) f(g(x)) = (x + 2)^2 - (x + 2) = x^2 + 3x + 2 - x - 2 = x^2 + 2x

(a(ii)) f(g(3)) = f(5) = 5^2 - 5 = 20

(b(i)) Tree diagram with correct probabilities on all branches (Red: 5/8, Blue: 3/8 for each pick) · must show Branch 1: Red (5/8), Blue (3/8); Branches splitting identically from each outcome.

(b(ii)) (5/8) × (5/8) = 25/64

(b(iii)) (5/8) × (3/8) + (3/8) × (5/8) = 30/64 = 15/32

Common mistakes: Mixing up the order of function composition (g(f(x)) vs f(g(x))); Forgetting to multiply probabilities on tree branches; Not considering both possible orders for 'one red and one blue'

Show mark scheme and worked solution

Marking points

M1 (a(i)) · 1 mark · Substitutes into and begins expansion of .

A1 (a(i)) · 1 mark · Fully simplifies to .

B1 (a(ii)) · 1 mark · Correctly calculates . (follow-through)

B1 (b(i)) · 2 marks · Draws complete tree diagram with correct probabilities (first and second pick). One mark for correct first split, one for second split and correct labels.

M1 (b(ii)) · 1 mark · Multiplies the probabilities for two reds: .

M1 (b(iii)) · 1 mark · Lists both orders: red then blue, blue then red, and finds their probabilities.

A1 (b(iii)) · 1 mark · Adds both to get correct final probability or 0.46875.

Worked solution

a(i) To find the composite function :

Expanding , so:

a(ii)

b(i) The tree diagram should show:

First pick: Red (5/8), Blue (3/8).

Second pick (each branch): Red (5/8), Blue (3/8), because of replacement. There are four final branches.

b(ii) Probability of both reds: (5/8) × (5/8) = 25/64 ≈ 0.391

b(iii) Probability of one red, one blue (either order):