Question 2 (b)
Defective components from three shifts: total probability and Bayes' theorem
5 marks
- Time to write
- About 10 minutes
Question
Question 2 (b) · 5 marksA company makes electrical components in three non-overlapping work shifts. It is found that 40%, 35%, and 25% of the components come from shifts 1, 2, and 3, respectively. Also, 5%, 9%, and 7% of the components made in shifts 1, 2, and 3, respectively, turn out defective. Determine:
What percentage of all the components is defective?
If a defective component is found, what is the probability that it came from shift 3?
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- What…?Is…?Two asks in one question: answer both.
- DetermineFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Events and joint probabilities 2 marks 3 min
- 2 (i) Percentage defective 1½ marks 2 min
- 3 (ii) Bayes' theorem for shift 3 1½ marks 2 min
- Check: arithmetic, units (₹, %), final answer underlined 2 min
Try it first. About 10 minutes for 5 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Events. Let S₁, S₂, S₃ be the events that a component comes from shift 1, 2 or 3 (mutually exclusive and exhaustive), and D that it is defective. P(S₁) = 0.40, P(S₂) = 0.35, P(S₃) = 0.25; P(D | S₁) = 0.05, P(D | S₂) = 0.09, P(D | S₃) = 0.07. Joint probabilities follow from the multiplication theorem; the reverse probability P(S₃ | D) follows from the definition of conditional probability, P(S₃ | D) = P(S₃ ∩ D) / P(D), which is Bayes' theorem.
| Shift | P(Sᵢ) | P(D | Sᵢ) | P(Sᵢ ∩ D) | P(Sᵢ | D) |
|---|---|---|---|---|
| 1 | 0.4 | 0.05 | 0.02 | 0.2899 |
| 2 | 0.35 | 0.09 | 0.0315 | 0.4565 |
| 3 | 0.25 | 0.07 | 0.0175 | 0.2536 |
| Total | 1 | 0.069 | 1.0000 |
(i) Percentage of components defective
P(D) = 0.02 + 0.0315 + 0.0175 = 0.069, so 6.9% of all components are defective.
(ii) Probability that a defective came from shift 3
P(S₃ | D) = P(S₃) P(D | S₃) / P(D) = 0.0175 / 0.069 = 0.2536.
About a quarter of the defective components come from shift 3; shift 2, with the highest defect rate, accounts for the largest share (0.4565).
Where the 5 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Events and joint probabilities | 2 |
| 2 | (i) Percentage defective | 1.5 |
| 3 | (ii) Bayes' theorem for shift 3 | 1.5 |
| Total | 5 |
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