Question 2 OR (e)
Binomial probabilities of a salesman's sales
6 marks
- Time to write
- About 12 minutes
Question
Question 2 OR (e) · 6 marksA salesman makes a sale to 65% of all the customers he contacts, on average. If 5 customers are contacted on one day, what is the probability that:
he sells to exactly 3 customers,
he could sell to at least one customer.
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
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Binomial model and parameters 2 marks 3 min
- 2 (i) Exactly three sales 2 marks 3 min
- 3 (ii) At least one sale 2 marks 3 min
- Check: arithmetic, units (₹, %), final answer underlined 2 min
Try it first. About 12 minutes for 6 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 distribution. Each of the 5 customers either buys (success) or does not, with the same chance p = 0.65 of a sale and independent decisions, so the number X of sales is binomial, X ~ B(5, 0.65), q = 0.35: P(X = x) = ⁵Cₓ pˣ q⁵⁻ˣ.
P(X = 3) = ⁵C₃ (0.65)³ (0.35)² = 10 × 0.274625 × 0.1225 = 0.3364.
P(X ≥ 1) = 1 − P(X = 0) = 1 − (0.35)⁵ = 1 − 0.0053 = 0.9947.
Comment. Three sales is the single most likely outcome of a day's five calls (about one day in three), and the salesman almost never ends a day of five calls without a sale.
Where the 6 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Binomial model and parameters | 2 |
| 2 | (i) Exactly three sales | 2 |
| 3 | (ii) At least one sale | 2 |
| Total | 6 |
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