Handling Data 7
4 marks · NormalThe probability that a machine produces a defective part is 0.1. The probability that a randomly chosen part was produced by machine A is 0.6, and by machine B is 0.4. The events are independent.
A part is selected at random. The possible outcomes are that the part was produced by machine A or B, and that the part is defective or not defective.
Construct a tree diagram to show all possible outcomes and their probabilities. Complete the probability table below.
| Produced by | Defective? | Probability |
|---|---|---|
| A | Yes | |
| A | No | |
| B | Yes | |
| B | No |
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
Must show: All probabilities filled in correctly; sum = 1
Common mistakes: Multiplying wrong way round; Forgetting to multiply stage probabilities
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Writes the correct probability for A and B at the first stage.
M2 · 1 mark · Writes the probabilities of 'defective' and 'not defective' for either A or B (0.1 and 0.9), showing independence.
A1 · 1 mark · Correctly fills in one full branch probability in the table (e.g., A & Yes = 0.06).
A2 · 1 mark · All probabilities filled in accurately, all branches, and the probabilities sum to 1.
Worked solution
First, draw the first stage with two branches: Machine A (0.6), Machine B (0.4).
For each machine, there are two branches: defective (0.1), not defective (0.9).
Multiply along each branch for the joint probability:
| Produced by | Defective? | Probability |
|---|---|---|
| A | Yes | 0.06 |
| A | No | 0.54 |
| B | Yes | 0.04 |
| B | No | 0.36 |
Sum: 0.06 + 0.54 + 0.04 + 0.36 = 1.0.