Part 1 · Probability & Statistics

1.3 Probability distributions

Binomial, Poisson and normal — a table of “which outcomes are how likely”

10-minute readPrerequisites: 1.1 What is probability?, 0.2 Reading math notation▶ Video (about 5 min)

What you will learn

Explain a probability distribution as “a list of possible outcomes and how likely each one is”
Explain how the binomial, Poisson and normal distributions differ, and when to use each
Decide which distribution fits test successes, accident or intervention counts, and sensor errors

Build intuition

Suppose you test a pedestrian-detection function 10 times on the same scene in which a pedestrian appears. Assume the detector’s true performance is “detects the pedestrian with 90% probability each time”. How many of the 10 tests will it pass?

It is tempting to say “9”, but the actual result varies from run to run. Listing the possible outcomes with their probabilities gives the following:

The most likely result is 9 detections (about 39%); passing all 10 tests happens only about 35% of the time.

A list of “possible outcomes” and “how likely each one is” like this is called a probability distribution. Even a detector with 90% true performance passes all 10 tests only about one time in three. Conversely, a perfect 10-out-of-10 does not mean its true performance is 100% (the same idea as in 1.8).

There are many kinds of probability distributions, but these three are the ones used most on this site:

Binomial distribution: how many successes in a fixed number of trials
Poisson distribution: how many times a rare event happens within a given distance or time
Normal distribution: how measurements scatter around their mean

Check with math

① Binomial distribution: how many successes out of n trials?

If you repeat a trial with success probability $p$ a total of $n$ times, the probability that the number of successes $X$ is exactly $k$ is given below. Here we assume that the trials do not influence each other and that the probability $p$ is the same every time.

$$P(X = k) = \binom{n}{k}\, p^k (1-p)^{n-k}$$

$p^k$ is the probability of “k successes”, $(1-p)^{n-k}$ is the probability that “the rest fail”, and $\binom{n}{k}$ (the number of ways to choose k out of n) counts the possible orders — which trials are the successes. For example, 9 successes out of 10 can happen in 10 different orders, depending on which trial is the single failure.

$$P(X = 9) = 10 \times 0.9^9 \times 0.1 \approx 0.387$$

The average number of successes is $np = 10 \times 0.9 = 9$.

② Poisson distribution: how many times does a rare event happen?

Now consider a very large number of trials $n$ with a very small probability $p$ per trial — for example, counting over tens of thousands of kilometres whether an accident happens in each kilometre. In this case the binomial distribution approaches a shape determined only by the average count $m = np$. This is the Poisson distribution.

$$P(N = k) = \frac{m^k e^{-m}}{k!}$$

$k!$ (k factorial) means $k \times (k-1) \times \cdots \times 1$, and by definition $0! = 1$. Setting k = 0 gives $P(N=0) = e^{-m}$ — the formula used in 1.8 as “the probability of finishing with zero accidents” (with m = accident rate λ × distance d).

Example: suppose a test vehicle experiences two interventions per day on average, where a safety driver takes over control (m = 2). The distribution of the daily number of interventions is:

Interventions per day	0	1	2	3	4	5 or more
Probability	13.5%	27.1%	27.1%	18.0%	9.0%	5.3%

Even at an average of two per day, roughly one day in seven has no interventions at all. That is why “no interventions today” alone does not show that the system has improved.

The Poisson distribution has three properties that safety calculations use again and again.

Rates add: if the rates (per km) of different causes are independent of each other, the overall rate is simply their sum. Adding up the rates of individual triggering conditions to get the overall rate (5.2) relies on this.
The scatter is √mean: the variance of a Poisson distribution equals its mean m, so the standard deviation is $\sqrt{m}$. With an average of 10 near misses a month, a swing of ±3 (≈ √10) is normal even when nothing has changed. That is why the alarm line in field monitoring is drawn at “mean + 3√mean” (5.11).
Waiting time until the next event: if the rate $\lambda$ is constant, the probability of no event during a time t is $e^{-\lambda t}$ (the waiting time follows the exponential distribution). The failure-rate unit FIT used in functional safety (one failure per billion hours) is such a rate: a 10 FIT component fitted to a million vehicles for a year (about 1,000 hours of driving) gives an expected 10⁻⁸ × 10⁶ × 10³ = 10 failures (4.7).

③ Normal distribution: the bell curve of measurement scatter

Scatter that arises from many small factors adding up — such as measurement error — often forms a symmetric bell shape centred on the mean $\mu$ (mu). This is the normal distribution (Gaussian distribution). Its shape is determined by just two numbers: the mean $\mu$ and the standard deviation $\sigma$ (the size of the spread, covered in detail in 1.4).

$$f(x) = \frac{1}{\sqrt{2\pi}\,\sigma} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$

You do not need to memorize this formula. For continuous values such as lengths (rather than counts), probability is represented by “the area under the curve”. The three numbers worth remembering are shown in the figure below.

The dark area is the range within ±σ of the mean (about 68%); including the light area gives the range within ±2σ (about 95%). Only about 0.27% falls outside ±3σ.

Example: suppose a LiDAR’s distance-measurement error follows a normal distribution with mean 0 and standard deviation σ = 2 cm. Then 95% of measurements fall within ±4 cm, and only about 0.27% fall outside ±6 cm (3σ). But 0.27% is not a small number in practice: a sensor that takes 100,000 measurements per second would exceed ±6 cm about 270 times every second.

④ Which distribution to use

Distribution	What it describes	What determines its shape	Example in automated driving
Binomial	The number of successes (or failures) in a fixed number of trials	Number of trials n, success probability p	How many of 100 test runs detected the pedestrian
Poisson	The number of rare events within a given distance or time	Average count m (= rate λ × distance or time)	Accidents or driver interventions per 10,000 km
Normal	The scatter of measurements caused by many small factors	Mean μ, standard deviation σ	Distance-measurement error of a sensor

Summary

A probability distribution is a list of “possible outcomes and how likely each one is”
For counts, use the binomial distribution when the number of trials is fixed, and the Poisson distribution when counting rare events over a distance or time. For the scatter of continuous measurements, use the normal distribution
Normal-distribution rule of thumb: about 68% within ±σ of the mean, about 95% within ±2σ, about 99.7% within ±3σ

Quick quiz

Q1. With a 90% detection rate, what is the probability of passing both of two tests?

81%. 0.9 × 0.9 = 0.81 (the binomial case with n = 2, k = 2).

Q2. A test vehicle averages two interventions per day. What is the probability of a day with zero interventions?

About 13.5%. With the Poisson distribution, $P(N=0) = e^{-2} \approx 0.135$.

Q3. If the error follows a normal distribution with mean 0 and standard deviation 2 cm, roughly what percentage of measurements exceed ±4 cm?

About 5%. ±4 cm corresponds to ±2σ, which contains about 95% of measurements, so about 5% fall outside.

Q4. Which of these is most naturally described by the Poisson distribution? (a) the number of successful detections in 100 tests (b) the number of accidents per 10,000 km (c) the error of a distance sensor

(b). Counting rare events within a given distance calls for the Poisson distribution. (a) has a fixed number of trials, so the binomial distribution fits; (c) is the scatter of continuous measurements, so the normal distribution is the usual starting point.