---
title: "Poisson call arrival probabilities"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:poisson-call-arrivals"
source: https://www.bctengineer.com/q/semester_8/telecommunications/poisson-call-arrivals
site: BCT Engineering
---

# Poisson call arrival probabilities

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2074 Chaitra, 8 marks
- Video: [What is a Poisson Process?](https://www.youtube.com/watch?v=5uLxFmhYnl0) — Iain Explains Signals, Systems, and Digital Comms
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/poisson-call-arrivals
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

On average, one call arrives every 5 seconds. During a period of 10 seconds, what is the probability that: i) No call arrives? ii) One call arrives? iii) Two calls arrive? iv) More than two calls arrive?

## Short answer

Arrival rate λ=0.2 calls/s, interval t=10 s, mean λt=2. Poisson probabilities: no call P(0)=e^{-2}=0.1353; one call P(1)=0.2707; two calls P(2)=0.2707; more than two calls P(>2)=1−(0.1353+0.2707+0.2707)=0.3233.

## Answer

Random call arrivals follow the **Poisson distribution**: the probability of exactly k calls in time t is
$$
P(k) = \frac{(\lambda t)^k}{k!}\,e^{-\lambda t}
$$

**Given:** one call every 5 s on average, so $\lambda = 0.2$ calls/s; t = 10 s.
$$
\lambda t = 0.2 \times 10 = 2
$$

#### (i) No call
$$
P(0) = e^{-2} = 0.1353
$$

#### (ii) One call
$$
P(1) = \frac{2^1}{1!}e^{-2} = 2 \times 0.1353 = 0.2707
$$

#### (iii) Two calls
$$
P(2) = \frac{2^2}{2!}e^{-2} = 2 \times 0.1353 = 0.2707
$$

#### (iv) More than two calls
$$
\begin{aligned}
P(k>2) &= 1 - [P(0) + P(1) + P(2)] \\
&= 1 - (0.1353 + 0.2707 + 0.2707) \\
&= 1 - 0.6767 = 0.3233
\end{aligned}
$$

| k | P(k) |
|---|---|
| 0 | 0.1353 |
| 1 | 0.2707 |
| 2 | 0.2707 |
| > 2 | 0.3233 |

**Answer:** (i) **0.1353**, (ii) **0.2707**, (iii) **0.2707**, (iv) **0.3233**.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What gives probability of exactly k calls in time t?
   A. lambda^k / k!
   B. e^(-lambda t) / k
   C. (1 - e^(-lambda t))^k
   D. (lambda t)^k e^(-lambda t) / k!
   Answer: D. Poisson distribution with mean lambda t. (Key point: Poisson formula)

2. One call per 5 s, t = 10 s. What is lambda t?
   A. 2
   B. 0.2
   C. 50
   D. 5
   Answer: A. 0.2 x 10 = 2. (Key point: Mean)

3. Probability of no call?
   A. 0.2707
   B. 0.1353
   C. 0.3233
   D. 0.8647
   Answer: B. e^-2 = 0.1353. (Key point: P(0))

4. How is P(k>2) found?
   A. P(2) = 0.2707
   B. 1 - P(0) = 0.8647
   C. 1 - [P(0)+P(1)+P(2)] = 0.3233
   D. P(0)+P(1)+P(2)
   Answer: C. Complement of the first three terms. (Key point: More than two)

## Related in Telecommunications

- [Traffic intensity, GOS and call duration probability](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atraffic-intensity-call-duration-probability) — Telecommunications, chapter 6
- [Traffic numerical: 600 calls, 40 lost, 120 s](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-600-calls-120s) — Telecommunications, chapter 6, asked 5×
- [GOS numerical: 1400 calls, 14 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-busy-hour-1400-calls) — Telecommunications, chapter 6, asked 5×
- [Traffic numerical: 1000 calls, 5 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atraffic-1000-calls-5-lost) — Telecommunications, chapter 6, asked 6×
- [GOS vs blocking; 600 calls, 2.5 min](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-blocking-600-calls-2-5min) — Telecommunications, chapter 6, asked 6×
- [Device holding time and calls per hour](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adevice-holding-time-calls) — Telecommunications, chapter 6

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- [RE: no more than 3 a's](https://www.bctengineer.com/question.md?key=semester_3%2Ftheory_of_computation%3Are-at-most-3-as) — Theory of Computation, chapter 2
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- [GSM frame structure and channels during a call](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Agsm-frame-call-channels) — Wireless Communication, chapter 8