Handling Data 7
4 marks · HardA company has two machines, A and B, producing light bulbs. - Machine A produces 60% of the bulbs, and machine B produces 40%. - 3% of the bulbs produced by machine A are defective. - 5% of the bulbs produced by machine B are defective. A bulb is selected at random. Draw a tree diagram showing all possible outcomes and use it to calculate the probability that a randomly chosen bulb is defective.
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
0.038 (±0.001)
Must show: Tree diagram must show branching for both machines and defective/non-defective outcomes with correct probabilities.
Also accept: 0.038; 0.0380; 3.8%
Common mistakes: 0.2; 0.035; 0.034
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Tree diagram structure with all branches and stages (Machine A/B, defective/not).
M2 · 1 mark · Correct probability multiplications along each branch (e.g. A & defective: ).
A1 · 1 mark · Accurately finds the defective probability from each machine: and . (follow-through)
A2 · 1 mark · Final total probability correct: . (follow-through)
Worked solution
Probability (A and defect) = 0.6 × 0.03 = 0.018 Probability (B and defect) = 0.4 × 0.05 = 0.02 Total defective probability = 0.018 + 0.02 = 0.038