Unit 8
8 marks · NormalThis question involves composite functions and probability tree diagrams.
Let the function be defined by and the function by .
Write down an expression for the composite function and simplify your answer.
Calculate the value of .
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.
Draw a tree diagram showing all possibile outcomes and label the probabilities on each branch.
Work out the probability that both balls chosen are red.
Work out the probability that one ball is red and one is blue (in any order).
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):