Probability of a Car Passing By: Poisson Process and Inter-Arrival Times
“Cars pass a point on a road independently. The probability of seeing a car in any 5-minute interval is 0.8. What’s the probability of seeing at least one car in a 10-minute interval?” This is a classic Poisson-process probability question that appears at quant trading firms (Jane Street, SIG, Optiver), data-science interviews, and ML interviews. The trick is recognizing the Poisson-process structure and applying the right formula. This guide covers the standard approach, the underlying theory, and the variations interviewers ask as follow-ups.
Problem Statement
The probability of seeing at least one car in any 5-minute interval is 0.8.
What’s the probability of seeing at least one car in a 10-minute interval?
The Naive (Wrong) Answer: 0.8 × 2 = 1.6
Probabilities don’t add like that. Probability is bounded above by 1; you can’t get 1.6.
The Correct Approach
Step 1: From the 5-minute probability, compute the probability of seeing zero cars in 5 minutes.
P(no car in 5 min) = 1 – 0.8 = 0.2
Step 2: Assume independence between non-overlapping intervals (Poisson process property). Then:
P(no car in 10 min) = P(no car in first 5 min) × P(no car in next 5 min) = 0.2 × 0.2 = 0.04
Step 3: The complement gives the answer.
P(at least one car in 10 min) = 1 – 0.04 = 0.96
The Underlying Math: Poisson Process
A Poisson process with rate λ has the property that events in disjoint intervals are independent, and the count in an interval of length t is Poisson distributed with parameter λt:
P(N(t) = k) = (λt)^k × e^(-λt) / k!
For “at least one event”:
P(N(t) ≥ 1) = 1 – P(N(t) = 0) = 1 – e^(-λt)
For our problem, given P(N(5) ≥ 1) = 0.8:
0.8 = 1 – e^(-5λ) → e^(-5λ) = 0.2 → λ = -ln(0.2) / 5 ≈ 0.322 cars per minute
For a 10-minute interval:
P(N(10) ≥ 1) = 1 – e^(-10λ) = 1 – e^(-3.22) ≈ 1 – 0.04 = 0.96
Same answer; explicit derivation. Strong candidates can derive λ from the given probability and apply it to other interval lengths.
Variations
Inter-arrival times
In a Poisson process, the time between consecutive events is exponentially distributed with rate λ. So if cars arrive at rate 0.322/min:
- Mean inter-arrival time: 1/λ ≈ 3.1 minutes
- P(next car arrives in less than 1 minute) = 1 – e^(-0.322) ≈ 0.275
- P(next car takes more than 5 minutes) = e^(-5 × 0.322) = e^(-1.61) ≈ 0.20
The “memoryless” property of exponential: how long you’ve already waited doesn’t affect the conditional waiting time for the next event.
Conditional probability of multiple events
“Given that you saw at least one car in 10 minutes, what’s the expected number?” Compute conditional expectation:
E[N(10) | N(10) ≥ 1] = E[N(10)] / P(N(10) ≥ 1) = (10λ) / 0.96 ≈ 3.35
(Approximately, for large λt where E[N(10) | N(10) ≥ 1] ≈ E[N(10)] / (1 – e^(-λt)).)
Multiple Poisson sources
Two Poisson processes with rates λ₁ and λ₂ combine into a Poisson process with rate λ₁ + λ₂. For example, two roads merging into one count point.
Thinning
If each event in a Poisson(λ) process is independently retained with probability p, the retained events form a Poisson(pλ) process. Used in modeling subset sampling.
Common Mistakes
- Adding probabilities directly. P(at least one in 10 min) ≠ 2 × P(at least one in 5 min). Probabilities of independent events compound multiplicatively, not additively.
- Forgetting the complement trick. The cleanest approach is “compute P(no events) and subtract from 1.” Direct computation of P(at least one) requires sum over all k ≥ 1, which is harder.
- Assuming independence when not warranted. “Cars passing” is a textbook Poisson assumption. Real traffic has dependencies (rush hour clusters, light cycles). For exam problems, the independence is usually given; for real-world modeling, verify it.
- Confusing rate units. λ has units of “events per unit time.” If λ = 0.322 per minute, then for hour-long intervals λt = 19.32. Watch the units.
- Using Poisson for non-Poisson processes. Poisson assumes independent increments and constant rate. If the rate varies (rush hour vs midnight), use a non-homogeneous Poisson process or partition into pieces with constant rate.
Frequently Asked Questions
What’s the expected interview answer?
0.96. Walk through: P(no car in 5 min) = 0.2; P(no car in 10 min) = 0.04 by independence; P(at least one car in 10 min) = 0.96. Mention the underlying Poisson-process assumption (events in disjoint intervals are independent). Strong candidates also derive λ explicitly to handle other interval lengths.
Why is the complement trick so useful?
P(at least one) is a sum over many cases (1, 2, 3, … events). P(zero) is a single computation. Computing the complement is almost always easier. This pattern shows up everywhere in probability — recognize it.
How does the Poisson-process structure help?
It guarantees independence between non-overlapping intervals, which lets you multiply probabilities. Without independence, the answer depends on correlation structure, which is usually unknown. The Poisson assumption is the simplest non-trivial model that gives a clean answer.
What if events are NOT independent?
The math gets harder. If events cluster (more likely after another), use Hawkes processes or other self-exciting models. If events anti-cluster (less likely after another), use renewal processes with non-exponential inter-arrival distributions. For interview purposes, independence is usually assumed; ask if unsure.
How does this generalize to real traffic engineering?
Traffic engineers use queueing theory built on Poisson arrivals at simple intersections. For complex networks, simulation models (microscopic traffic simulators) capture dependencies that pure Poisson misses. The interview question is the foundational case; real-world models layer on complications.
See also: Probability Distribution Functions • Expected Value • Random Walks and Stopping Times