Question 2 OR (e)
Poisson probabilities of accidents on a motorway stretch
9 marks
- Time to write
- About 18 minutes
Question
Question 2 OR (e) · 9 marksAn insurance company is studying the risk of accidents on a particular stretch of motorway so as to adjust its premium rates. Accidents occur on this stretch at the rate of 4 per month. Find the probability that in a given month there will be:
not a single accident,
more than 2,
at least 1 accident.
(Given e^-4 = 0.01832)
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
- at least 1The question sets a minimum: give 1 or more.
- FindFormula 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 Poisson model and formula 2 marks 4 min
- 2 P(0), P(1), P(2) 3 marks 5 min
- 3 (i) P no accident 1 mark 2 min
- 4 (ii) P more than two 2 marks 3 min
- 5 (iii) P at least one 1 mark 2 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 18 minutes for 9 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. Accidents on a stretch of road are rare, separate events occurring at random over time, with a steady average of 4 a month; the number X of accidents in a month is therefore taken as a Poisson variable with m = 4. Its probabilities are P(X = x) = e⁻ᵐ mˣ / x!, for x = 0, 1, 2, …, and the mean and the variance both equal m. Counts of rare events over time, such as accidents, are the typical setting for this distribution.
Probabilities needed (e⁻⁴ = 0.01832)
| x | Working | P(X = x) |
|---|---|---|
| 0 | e⁻⁴ × 4⁰ / 0! = 0.01832 | 0.01832 |
| 1 | e⁻⁴ × 4¹ / 1! = 0.01832 × 4 | 0.07328 |
| 2 | e⁻⁴ × 4² / 2! = 0.01832 × 8 | 0.14656 |
| 0 to 2 | 0.23816 |
P(X = 0) = 0.01832: about a 1.8% chance of an accident-free month.
P(X > 2) = 1 − [P(0) + P(1) + P(2)] = 1 − 0.23816 = 0.76184.
P(X ≥ 1) = 1 − P(0) = 1 − 0.01832 = 0.98168.
Comment for the insurer. In about three months out of four the stretch will see more than two accidents, and a month with no accident at all is very rare, so premiums for vehicles using this stretch should reflect a high claim rate.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Poisson model and formula | 2 |
| 2 | P(0), P(1), P(2) | 3 |
| 3 | (i) P(no accident) | 1 |
| 4 | (ii) P(more than two) | 2 |
| 5 | (iii) P(at least one) | 1 |
| Total | 9 |
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