How do you read a probability tree?

From the root rightwards. Each level is one stage of the experiment, each branch one possible outcome of that stage, and the number on a branch is a conditional probability given everything that happened before it. The probability of a whole path comes from multiplying every number along it, and the probability of an event from adding the paths that realise it.

Why do you multiply along a branch but add across paths?

Because these are two different rules with two different reasons. Multiplying along a branch is the multiplication rule, that is the definition of conditional probability written the other way round. Adding across paths is legitimate because different paths are disjoint events — they exclude one another, so the sum counts nothing twice.

When may you write P(A ∪ B) = P(A) + P(B)?

Only when the events are disjoint. In general the overlap would be counted twice — once in P(A) and once in P(B) — so it has to be subtracted and the formula reads P(A) + P(B) − P(A ∩ B). A result larger than one is the usual sign that something has gone wrong.

What is the Bernoulli scheme?

A run of n trials satisfying three conditions at once: each has exactly two outcomes called success and failure, the probability of success p is the same in every trial, and the trials are independent. Coin tosses, draws with replacement and successive units from a very large batch fit that description; drawing without replacement from a small set does not, because p changes after every step.

How do you compute the probability of at least one success?

Through the complement, not by summing the cases from 1 to n. Failing in every one of the n trials has probability q to the n, so the value wanted is 1 minus that power. It is one line of arithmetic instead of n terms, and it is exactly the shortcut the classical-probability lesson introduced with the complementary event.

Must a tree carry the same probabilities at every stage?

No. The numbers on the branches are conditional probabilities, so they may differ between levels and between nodes of the same level — which is what happens when drawing without replacement. The Bernoulli scheme is the special case where they are constant, and that is precisely why it can be written as one formula instead of drawn as a tree.

Advanced level

Probability trees and the Bernoulli scheme

Q: What is the difference between disjoint and independent events?

Disjoint events cannot happen together: P(A ∩ B) = 0. Independent events can, it is just that neither affects the other: P(A ∩ B) = P(A) times P(B). The two ideas are not merely different but opposed — two disjoint events of non-zero probability are certainly not independent, because one occurring drops the other’s chance to zero.

Q: Where does the binomial coefficient in the Bernoulli formula come from?

From the number of paths. Every path with exactly k successes carries the same probability p to the k times q to the n−k, because it multiplies the same factors in a different order. There are as many such paths as there are ways of choosing k of the n positions for the successes — and that is the number of combinations from the combinatorics lesson.

A multi-stage experiment draws as a tree: probabilities multiply along the branches and add across the paths. That one picture holds the formula for the union of two events, the difference between disjoint and independent events, and the whole Bernoulli scheme — including where its binomial coefficient comes from.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Where this is used

Real situations where you count exactly the way this lesson teaches:

Inspecting a batch of goods
From a batch in which 5 per cent of units are faulty, a storekeeper pulls 10 for inspection. The probability that none of them turns out faulty is 0.95 to the tenth power, that is 0.599 — so almost 40 per cent of such inspections catch at least one defect. Had 20 per cent been faulty, a sample of ten would pass the batch only 11 per cent of the time, and that is the whole mathematics behind choosing a sample size.
Guessing a multiple-choice test
A test has 10 questions, each with four options and one correct answer. Guessing is a Bernoulli scheme with p equal to 1/4, so the probability of not a single hit is 0.75 to the tenth power, about 5.6 per cent. The probability of passing with a threshold of 6 correct answers, on the other hand, is only 0.0197 — fewer than 2 cases in 100.
Redundancy in a server room
Three independent servers, each of which fails in a given month with probability 0.02, all go down together with probability 0.02 cubed, that is 0.000008. The service therefore runs 99.9992 per cent of the time, while a single server manages 98 per cent. Two extra machines cut annual downtime from about 175 hours to under 5 minutes.
Overlapping groups in a survey
Of 200 respondents, 90 read the newsletter and 60 follow the social media profile, and 25 people do both. At least one channel therefore reaches 90 + 60 − 25 = 125 people, that is 62.5 per cent of the group — not the 75 per cent a plain addition would give. Those 25 people are exactly the term the union formula subtracts.

All formulas

Multiplying along a branch
$P(\text{path}) = p_1 \cdot p_2 \cdot \ldots \cdot p_n$
the probability of one path of the tree
Adding across paths
$P(A) = \sum_{\text{paths} \subset A} P(\text{path})$
an event is a set of paths, so they add
Union of events
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
the overlap is counted twice, so subtracted once
Disjoint events
$A \cap B = \varnothing \implies P(A \cup B) = P(A) + P(B)$
nothing to subtract
Bernoulli scheme
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
k successes in n independent trials
At least one success
$P(X \geq 1) = 1 - (1-p)^n$
through the complement, with no summing

The previous lesson drew trees to show total probability and Bayes’ theorem. The picture did the work, but its rules were never named. We name them now — and then find that one particular tree recurs so often that it has earned a formula of its own.

The two rules of a tree

A multi-stage experiment is recorded as a probability tree: a level is a stage, a branch is a possible outcome of that stage, and the number on a branch is the probability of that outcome given what has already happened.

Two rules govern it, and it is worth knowing where each comes from:

multiply along a branch — this is the multiplication rule $P(A \cap B) = P(B) \cdot P(A|B)$ of the previous lesson;
add across paths — legitimate because different paths exclude one another, so the sum counts nothing twice.

Two draws without replacement from an urn holding 3 white balls and 2 black. The numbers on the second level differ between the two nodes — because the urn looks different after the first draw. The four outcomes add up to 20/20, that is to one.

Two balls are drawn without replacement from an urn with 3 white balls and 2 black. What is the probability of drawing exactly one white?

Disjoint is not independent

This is the pair of ideas confused most often, and the difference is not subtle — they actually rule each other out.

	disjoint	independent
meaning	cannot happen together	one does not affect the other
formula	$P(A \cap B) = 0$	$P(A \cap B) = P(A) \cdot P(B)$
in a union	$P(A) + P(B)$	subtract $P(A) \cdot P(B)$

Take a die roll, with $A$ — an even number — and $B$ — an odd one. The events are disjoint, since $A \cap B = \varnothing$ . Are they independent? Check:

P(A \cap B) = 0, \qquad P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

They are not. Nor is this an accident of this die: two disjoint events of non-zero probability are never independent — because learning that $A$ occurred drops the chance of $B$ to zero, and that is the largest change there is.

Independence does turn up between different experiments or different stages: two coin tosses, two draws with replacement, the failures of two separate servers.

The union of two events

The event $A \cup B$ occurs when $A$ occurs, when $B$ occurs, or when both do. Plain addition $P(A) + P(B)$ counts the overlap twice, so it has to be subtracted once:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

For disjoint events the last term vanishes and only the addition is left — exactly what the rule about adding across paths rests on.

One card is drawn from a 52-card deck. What is the probability that it is a heart or a king?

The Bernoulli scheme

The tree drawn most often is the one whose levels all look the same. We speak of a Bernoulli scheme when three conditions hold at once:

every trial has exactly two outcomes — success and failure;
the probability of success $p$ is the same in every trial;
the trials are independent.

Coin tosses, draws with replacement and successive units from a very large batch fit that description. Drawing without replacement from a small set — as in the urn tree above — does not, because $p$ changes after every step.

Three coin tosses. All eight paths carry the same probability 1/8, because each multiplies three identical factors. Three of them show exactly two heads — as many as there are ways of choosing two positions out of three.

The drawing contains the whole formula. Every path with $k$ successes multiplies $p$ exactly $k$ times and $q = 1-p$ exactly $n-k$ times, so it carries probability $p^k q^{n-k}$ — whatever the order. There are as many such paths as there are ways of choosing $k$ positions out of $n$ , that is $\binom{n}{k}$ , the combinations of the counting lesson. Hence

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

A die is rolled 5 times. What is the probability that a six comes up exactly twice?

At least one success

The question “will it happen even once” comes up so often in this scheme that a shortcut is worth having. Summing from $k = 1$ to $k = n$ is the wrong method here, not merely the slow one: the complement is a single term.

P(X \geq 1) = 1 - P(X = 0) = 1 - (1-p)^n

A batch contains 5% faulty units. Ten units are drawn. What is the probability of hitting at least one faulty one?

Exercises

The set asks three things, in the order of the lesson. The $P(A \cup B)$ prompt gives $P(A)$ , $P(B)$ and $P(A \cap B)$ — the union formula has to be assembled, subtraction included. The $P(A \cap B)$ prompt gives $P(A)$ and $P(B|A)$ , that is two branches of one path, and asks for their product. The $P(X = k)$ prompt gives the number of trials $n$ and the success probability $p$ , so it asks for the Bernoulli formula directly.

In every case the answer is a fraction in lowest terms.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

P(A∪B): P(A) = 7/10, P(B) = 3/5, P(A∩B) = 1/2

The generator does not ask for a “disjoint or independent” verdict — the answer would be a word rather than a number. That distinction is in the table and in the die counterexample above, and it is worth returning to, because it recurs in every problem about a union of events.

Common mistakes

Confusing disjointness with independence — the first says “not together”, the second “no influence”. Disjoint events of non-zero probability are never independent.
Adding $P(A) + P(B)$ for events that can happen together — the overlap gets counted twice. A result larger than one is the commonest symptom.
Multiplying the probabilities of stages that are not independent — on a tree the second number is conditional; when drawing without replacement it differs at every node.
Dropping the binomial coefficient from the Bernoulli formula — $p^k q^{n-k}$ is the probability of one path, and there are $\binom{n}{k}$ of them.
Applying the Bernoulli scheme to drawing without replacement — there $p$ is not constant and the trials are not independent, so the formula simply does not hold.
Computing “at least one” by summing every $k \geq 1$ — one calculation on the complement gives the same answer and has nothing in it to lose.

Formula card

Topic: Trees and the Bernoulli scheme

Multiplying along a branch
$P(\text{path}) = p_1 \cdot p_2 \cdot \ldots \cdot p_n$
the probability of one path of the tree
Adding across paths
$P(A) = \sum_{\text{paths} \subset A} P(\text{path})$
an event is a set of paths, so they add
Union of events
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
the overlap is counted twice, so subtracted once
Disjoint events
$A \cap B = \varnothing \implies P(A \cup B) = P(A) + P(B)$
nothing to subtract
Bernoulli scheme
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
k successes in n independent trials
At least one success
$P(X \geq 1) = 1 - (1-p)^n$
through the complement, with no summing

Three coin tosses as a tree. Eight paths, each of probability 1/8 — because every one of them multiplies three branches worth 1/2. Exactly two heads happen on three of them, and that count is the binomial coefficient in the Bernoulli formula.

Frequently asked questions

Luminous histogram bars settling into the soft bell of a distribution, one outlying point brighter than the rest, against a cosmic nebula in violet and magenta

Statistics and probability

The average person does not exist

All of descriptive statistics rests on a single trade: you give up information about a crowd and receive one number in return. The trade is often worth making, but it is never free — and it pays to know exactly what gets lost in it. From the mortality tables of 1662, through Quetelet's average man, to the aircraft nobody ever saw.

16 min