Syllabus topics: Discrete distributions — Binomial, Poisson, Geometric, Negative Binomial: definitions, properties and examples. Continuous distributions — Uniform, Normal (Gaussian), Exponential, Gamma: definitions, properties and applications. Joint, marginal and conditional distributions. Introduction to the Central Limit Theorem.
THE BIG IDEA
A probability distribution is a ready-made pattern for how a certain kind of randomness behaves. Recognise which pattern fits your situation and every probability you need is a formula away, instead of counting outcomes by hand.
IN DEPTH
You could work out the chance of getting exactly 7 heads in 10 coin tosses by listing all 1,024 possible sequences and counting. Or you could notice that this is a fixed number of independent yes/no trials with a constant success probability — a binomial situation — and get the answer in one line.
The skill this unit tests is recognising the pattern. The formulas are on the formula sheet; knowing which one applies is what earns the marks.
| Situation | Distribution |
|---|---|
| Fixed number of yes/no trials, count successes | Binomial |
| Count rare events in a fixed interval of time or space | Poisson |
| Trials until the first success | Geometric |
| Trials until the rth success | Negative binomial |
| Every value in a range equally likely | Uniform |
| Measurements clustering around an average, bell-shaped | Normal |
| Waiting time until the next event | Exponential |
| Waiting time until the rth event | Gamma |
IN DEPTH
A basketball player makes 70% of her free throws. She takes 10. What is the chance she makes exactly 8? Each throw is independent, has the same success probability, and has only two outcomes. That is a binomial situation.
FORMULA
P(X = k) = ⁿCₖ · pᵏ · (1−p)ⁿ⁻ᵏ
In plain English:
| Formula | |
|---|---|
| Mean | μ = np |
| Variance | σ² = np(1−p) |
| SD | σ = √(np(1−p)) |
Worked example. n = 10, p = 0.7, find P(X = 8).
About a 23% chance. Mean = np = 7, so 8 is slightly above average — plausible, and the arithmetic agrees.
Excel: =BINOM.DIST(8, 10, 0.7, FALSE)
IN DEPTH
A call centre receives about 3 calls per minute. What is the chance of exactly 5 calls in the next minute? There is no fixed number of "trials" here — a call can arrive at any instant. You are counting rare events in a fixed interval. That is Poisson.
FORMULA
P(X = k) = e^(−λ) · λᵏ / k!
| Formula | |
|---|---|
| Mean | μ = λ |
| Variance | σ² = λ |
KEY INSIGHT
For a Poisson distribution the mean and the variance are equal. No other common distribution does this, so it is the signature to look for — and a standard two-mark question.
Worked example. λ = 3, find P(X = 5).
When n is large and p is small, with np = λ moderate, the binomial is very well approximated by the Poisson. Rule of thumb: n ≥ 20 and p ≤ 0.05.
| k | Binomial(1000, 0.003) | Poisson(3) |
|---|---|---|
| 0 | 0.049385 | 0.049787 |
| 1 | 0.148600 | 0.149361 |
| 2 | 0.223347 | 0.224042 |
| 3 | 0.223570 | 0.224042 |
Nearly identical — and the Poisson is far easier to compute, since ¹⁰⁰⁰C₃ is
unpleasant by hand. (These figures are computed in
03_random_variables_distributions.py.)
The number of trials until the first success.
FORMULA
P(X = k) = (1−p)^(k−1) · p
You fail k−1 times, then succeed.
| Formula | |
|---|---|
| Mean | μ = 1/p |
| Variance | σ² = (1−p)/p² |
WORKED EXAMPLE
Rolling a die until the first six: p = 1/6, so on average it takes 1/(1/6) = 6 rolls. That matches intuition, which is a good sign.
P(first six on the 3rd roll) = (5/6)² × (1/6) = 25/216 = 0.1157
Memorylessness: the geometric is the only discrete distribution with this property. Having failed ten times does not improve your chances on the eleventh — the die has no memory.
The number of trials until the rth success — the geometric generalised.
FORMULA
P(X = k) = ^(k−1)C_(r−1) · pʳ · (1−p)^(k−r)
| Formula | |
|---|---|
| Mean | μ = r/p |
| Variance | σ² = r(1−p)/p² |
When r = 1 this reduces exactly to the geometric distribution.
WORKED EXAMPLE
P(the 3rd six occurs on the 10th roll), p = 1/6: ⁹C₂ × (1/6)³ × (5/6)⁷ = 36 × 0.00463 × 0.2791 = 0.0465
Every value in [a, b] is equally likely.
FORMULA
f(x) = 1/(b−a) for a ≤ x ≤ b
| Formula | |
|---|---|
| Mean | μ = (a+b)/2 |
| Variance | σ² = (b−a)²/12 |
The density is a flat rectangle of height 1/(b−a) and width (b−a) — so its area is exactly 1, as every PDF's must be.
WORKED EXAMPLE
A bus arrives uniformly between 0 and 20 minutes from now. Expected wait = 10 minutes. P(waiting less than 5) = 5/20 = 0.25.
The most important distribution in statistics.
IN DEPTH
Measure the heights of a thousand students. A few are very short, a few very tall, and most cluster near the average, tapering off symmetrically in both directions. Plot it and you get the bell curve. Heights, weights, measurement errors, exam scores, blood pressure — an enormous range of natural quantities follow this shape, and Section 3.9 explains why.
FORMULA
f(x) = (1 / (σ√(2π))) · e^(−(x−μ)²/(2σ²))
You will rarely use this directly — you use tables or software. What you must know is its properties:
| Interval | Contains |
|---|---|
| μ ± 1σ | 68.27% of the data |
| μ ± 2σ | 95.45% |
| μ ± 3σ | 99.73% |
Worth committing to memory: it lets you sanity-check any normal calculation in seconds.
Every normal distribution can be converted to the standard normal (μ = 0, σ = 1):
FORMULA
z = (x − μ) / σ
A z-score says how many standard deviations from the mean a value lies. z = 2 means "two standard deviations above average", regardless of the original units. That is what makes different scales comparable.
WORKED EXAMPLE
IQ scores are Normal(100, 15).
Critical values worth memorising:
| Confidence | z |
|---|---|
| 90% | 1.645 |
| 95% | 1.96 |
| 99% | 2.576 |
Excel: =NORM.DIST(x, mean, sd, TRUE) for the CDF, =NORM.INV(p, mean, sd)
for the inverse.
The waiting time until the next event, when events occur at a constant average rate λ. It is the continuous partner of the Poisson: if events per hour are Poisson(λ), the gap between events is Exponential(λ).
FORMULA
f(x) = λe^(−λx) for x ≥ 0 F(x) = 1 − e^(−λx)
| Formula | |
|---|---|
| Mean | μ = 1/λ |
| Variance | σ² = 1/λ² |
WORKED EXAMPLE
Buses arrive at 0.5 per minute (λ = 0.5). Mean wait = 1/0.5 = 2 minutes. P(waiting more than 3 minutes) = e^(−0.5 × 3) = e^(−1.5) = 0.2231.
KEY INSIGHT
P(X > s + t | X > s) = P(X > t)
You have already waited 2 minutes. The probability of waiting 3 more is exactly the same as it was at the start. The bus does not "become due".
The exponential is the only continuous distribution with this property, and it is a favourite exam question. It is also why it models the lifetime of components that fail randomly rather than by wearing out — a fuse, not a tyre.
The waiting time until the rth event. The exponential generalised, exactly as the negative binomial generalises the geometric.
FORMULA
f(x) = (λ^α / Γ(α)) · x^(α−1) · e^(−λx) for x > 0
| Formula | |
|---|---|
| Mean | μ = α/λ |
| Variance | σ² = α/λ² |
Special cases worth naming:
The Γ function extends factorials to non-integers: Γ(n) = (n−1)! for positive integers, and Γ(1/2) = √π.
| One event | r events | |
|---|---|---|
| Discrete (count trials) | Geometric | Negative binomial |
| Continuous (measure time) | Exponential | Gamma |
When two random variables are considered together.
Joint distribution: p(x, y) = P(X = x and Y = y)
Marginal distribution: sum out the other variable — p(x) = Σ_y p(x, y). It is called "marginal" because it was traditionally written in the margins of the joint table.
Conditional distribution: p(y|x) = p(x, y) / p(x)
Independence: X and Y are independent when p(x, y) = p(x) × p(y) for every pair.
WORKED EXAMPLE
| Y=0 | Y=1 | Marginal of X | |
|---|---|---|---|
| X=0 | 0.2 | 0.3 | 0.5 |
| X=1 | 0.1 | 0.4 | 0.5 |
| Marginal of Y | 0.3 | 0.7 | 1.0 ✓ |
THE BIG IDEA
Take samples of reasonable size from any population, whatever its shape, and the distribution of the sample means will be approximately normal.
IN DEPTH
Roll one die: every outcome from 1 to 6 is equally likely — a flat, uniform distribution, nothing bell-shaped about it. Now roll thirty dice and record the average. Do that many times and plot the averages. You get a bell curve.
The population was uniform. The distribution of averages is normal. That is the Central Limit Theorem, and it is the reason the normal distribution appears everywhere in nature: most measurable quantities are the sum of many small independent influences.
FORMULA
For a population with mean μ and standard deviation σ, the sampling distribution of the mean x̄ for sample size n has:
Mean of x̄ = μ SD of x̄ = σ/√n — this is the standard error Shape → normal as n grows, whatever the population's shape
Rule of thumb: n ≥ 30 is enough for most populations. A population that is already normal gives a normal sampling distribution at any n.
KEY INSIGHT
It is what makes inference possible. You take one sample, and the CLT tells you how the mean of that sample behaves relative to the population mean — so you can build a confidence interval or run a hypothesis test. Every technique in Unit 5 rests on it.
Notice too that the standard error is σ/√n, not σ/n. To halve your uncertainty you need four times the data. That single fact governs the cost of every survey and experiment ever run.
Worked example. A population has μ = 100, σ = 20. For samples of n = 25:
PROBLEM 1
A factory's items are 5% defective. In a random sample of 20, find the probability that (a) exactly 2 are defective, (b) at most 1 is defective. Also find the mean and variance.
Solution. Binomial with n = 20, p = 0.05.
(a) P(X = 2) = ²⁰C₂ × 0.05² × 0.95¹⁸
(b) P(X ≤ 1) = P(0) + P(1)
Mean = np = 20 × 0.05 = 1.0; Variance = np(1−p) = 1 × 0.95 = 0.95
(Since n is largish and p small, Poisson(λ = 1) gives P(X=2) = 0.1839 — close to 0.1887, as expected.)
PROBLEM 2
Heights are normally distributed with mean 165 cm and standard deviation 8 cm. Find (a) P(height > 175), (b) P(160 < height < 170), (c) the height exceeded by only 5% of people.
Solution.
(a) z = (175 − 165)/8 = 1.25 P(Z > 1.25) = 1 − 0.8944 = 0.1056, about 10.6%
(b) z₁ = (160 − 165)/8 = −0.625; z₂ = (170 − 165)/8 = 0.625 P = 0.7340 − 0.2660 = 0.4680, about 46.8%
(c) The 95th percentile has z = 1.645 x = μ + zσ = 165 + 1.645(8) = 165 + 13.16 = 178.16 cm
PROBLEM 3
A website receives 4 visitors per minute on average. Find the probability of (a) exactly 6 visitors in a minute, (b) no visitors in a minute, (c) more than 2 visitors.
Solution. Poisson with λ = 4.
(a) P(X=6) = e^(−4) × 4⁶ / 6! = 0.018316 × 4096 / 720 = 0.1042
(b) P(X=0) = e^(−4) × 4⁰ / 0! = 0.0183
(c) P(X > 2) = 1 − [P(0) + P(1) + P(2)]
Using the complement is far quicker than summing P(3), P(4), P(5), … to infinity. Always look for that shortcut on "more than" questions.
Two marks
Five marks
Ten marks
Explain the discrete probability distributions — binomial, Poisson, geometric and negative binomial — with formulas, moments and examples.
Explain the continuous distributions — uniform, normal, exponential and gamma — with their properties and applications.
COMMON ERRORS