Probability distributions in practice

Random variables and distributions · examples

Probability distributions in practice

The same formula may describe free throws, calls or waiting times. The important step is recognising what is measured and which assumptions make the model plausible.

1. Start from the question, not the formula

One yes/no outcome?

A single trial suggests Bernoulli; the number of successes over several trials suggests binomial.

A count in time or space?

Poisson is a candidate when events occur independently at a stable average rate.

How long until something happens?

Geometric for a discrete number of attempts; exponential for continuous waiting time between Poisson events.

A continuous, symmetric measurement?

Normal may fit when many small independent effects add together, provided physical boundaries are not a problem.

A model is an approximation. Check support, independence, parameter stability and the data-generating mechanism. A histogram that merely resembles a curve is not enough.

2. Concrete discrete examples

Bernoulli

Component quality control

A randomly chosen component is defective with probability 3%. Let $X = 1$ for a defect and 0 otherwise.

X \sim Ber (0.03), P (X = 1) = 0.03

For the number of defective items among 100 independent components, use a binomial model instead.

Binomial

Basketball free throws

A player takes 10 independent shots and scores each with probability 0.7. The probability of exactly 8 scores is:

P (X = 8) = (\binom{10}{8}) {0.7}^{8} {0.3}^{2} \approx 0.2335

Fatigue or learning may violate the constant-probability and independence assumptions.

Geometric

Attempts until connection

Each attempt to connect to a service succeeds with probability 0.2. The probability that the first success occurs on attempt four is:

P (X = 4) = {0.8}^{3} \cdot 0.2 = 0.1024

This means three failures followed by one success, assuming independent attempts with constant probability.

Poisson

Calls to a switchboard

An average of 3 calls arrive every ten minutes. With a stable rate and independent arrivals, the probability of exactly 5 is:

P (X = 5) = e^{- 3} \frac{3^{5}}{5!} \approx 0.1008

If the rate changes through the day, split the model into time bands or use a non-homogeneous process.

3. Concrete continuous examples

Uniform

Waiting for a bus

A perfectly punctual bus comes every 12 minutes and you arrive at a random time, so $X \sim U (0, 12)$ .

P (X \leq 3) = \frac{3}{12} = 0.25

The model assumes the passenger's arrival is not synchronised with the timetable.

Exponential

Component lifetime

A component has a mean life of 1000 hours and constant failure rate $λ = 0.001$ per hour.

P (T > 500) = e^{- 0.5} \approx 0.6065

A constant rate means no ageing, which may be unrealistic for components that wear out.

Normal

Heights in a population

Suppose height is approximately normal with mean 175 cm and standard deviation 7 cm. The interval 168–182 cm is $μ \pm σ$ .

P (168 \leq X \leq 182) \approx 0.6827

The 68–95–99.7 rule applies to normal distributions, not automatically to every data set with a mean and standard deviation.

4. Lab: from a binomial model to experiments

Change the number of shots, success probability and observed count. Light bars show theoretical probabilities; after simulation, outlines show frequencies from 1000 series.

Shots

n

: 10

Probability

p

: 0.7

Observed successes

k

: 7

Expected value: 7
Theoretical probability of k: 26.68%
Simulated frequency of k: —

Press “Simulate 1000 series” to compare theory and experiment.