Question 2 (d)
Poisson probabilities of equipment failures in a week
5 marks
- Time to write
- About 10 minutes
Question
Question 2 (d) · 5 marksA manufacturing plant has two equipment failures a week on average. What is the probability of:
exactly three equipment failures during a given week,
two or more equipment failures within a given week,
no equipment failure at all in a given week?
[Given: e^-2 = 0.1353]
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
What the paper leaves open
As printed, the third item is numbered (ii) a second time. The question's average is written in words (two) and the exponent in the given value is a superscript.
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Poisson model and formula 1 mark 2 min
- 2 Exactly three failures 1½ marks 2 min
- 3 Two or more failures 1½ marks 2 min
- 4 No failure 1 mark 2 min
- Check: arithmetic, units (₹, %), final answer underlined 1 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.
The model. Breakdowns are rare, independent events occurring at random at a steady average of two a week, so the number X of failures in a week is a Poisson variable with m = 2: P(X = x) = e⁻ᵐ mˣ / x!, with mean and variance both equal to m. e⁻² = 0.1353. (The paper numbers the third item (ii) again; it is answered here as the third item.)
| x | Working | P(X = x) |
|---|---|---|
| 0 | e⁻² × 2⁰ / 0! | 0.1353 |
| 1 | e⁻² × 2¹ / 1! = 0.1353 × 2 | 0.2706 |
| 0 or 1 | 0.4059 | |
| 3 | e⁻² × 2³ / 3! = 0.1353 × 8 / 6 | 0.1804 |
Exactly three failures in a given week: P(X = 3) = 0.1804.
Two or more failures in a given week: P(X ≥ 2) = 1 − [P(0) + P(1)] = 1 − 0.4059 = 0.5941.
No failure at all in a given week: P(X = 0) = e⁻² = 0.1353.
Comment. The plant can expect a failure-free week only about one week in seven, and two or more failures in more than half of all weeks, which supports keeping maintenance staff and spares on hand.
Where the 5 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Poisson model and formula | 1 |
| 2 | Exactly three failures | 1.5 |
| 3 | Two or more failures | 1.5 |
| 4 | No failure | 1 |
| Total | 5 |
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