Probability rules, Venn diagrams and tree diagrams

Probability measures uncertainty on a scale from $0$ to $1$. An impossible event has probability $0$, a certain event has probability $1$, and larger values mean stronger evidence that the event will occur.

This lesson develops the rules needed to combine events correctly. The central challenge is not arithmetic. It is translating words such as or, and, not and independent into precise mathematical statements.

Prerequisites

You should be able to:

• calculate with fractions, decimals and percentages;
• list outcomes systematically and recognise equally likely outcomes;
• use basic set notation for unions, intersections and complements;
• multiply and add fractions without prematurely rounding.

Review probability foundations if sample spaces and relative frequency are unfamiliar.

Outcomes, sample spaces and events

A random experiment has an uncertain result. An outcome is one possible result. The sample space $S$ is the set of all possible outcomes, and an event is a subset of $S$.

For one roll of a fair six sided die,

$$S=\{1,2,3,4,5,6\}.$$

If $A$ is the event “an even number is rolled”, then

$$A=\{2,4,6\}.$$

Because the six outcomes are equally likely,

$$\operatorname{P}(A)=\frac{\text{number of outcomes in }A}{\text{number of outcomes in }S} =\frac36=\frac12.$$

This counting formula is valid only when the elementary outcomes are equally likely. For an unfair spinner, probabilities must come from the model or from data, not merely from counting sectors.

Every event $A$ satisfies

$$\boxed{0\leq \operatorname{P}(A)\leq1}, \qquad \operatorname{P}(S)=1, \qquad \operatorname{P}(\varnothing)=0.$$

Worked example 1: construct the sample space

A fair coin is tossed and a fair die is rolled. Find the probability of obtaining a head and a number greater than $4$.

There are $2\times6=12$ equally likely ordered outcomes:

$$S=\{H1,H2,\ldots,H6,T1,T2,\ldots,T6\}.$$

The required event is $\{H5,H6\}$, so

$$\operatorname{P}(\text{head and number greater than }4) =\frac2{12}=\boxed{\frac16}.$$

The word “and” requires both conditions to hold in the same outcome.

Self check 1

Two fair four sided dice, numbered $1$ to $4$, are rolled. Find the probability that their total is $5$.

Answer

There are $4^2=16$ equally likely ordered pairs. The successful pairs are

$$(1,4),(2,3),(3,2),(4,1).$$

Therefore

$$\boxed{\operatorname{P}(\text{total }5)=\frac4{16}=\frac14}.$$

The pairs $(1,4)$ and $(4,1)$ are different outcomes.

Complements: the probability of not happening

The complement $A'$ contains every outcome in $S$ that is not in $A$. Exactly one of $A$ and $A'$ must occur, so

$$\boxed{\operatorname{P}(A')=1-\operatorname{P}(A)}.$$

Complements are especially useful for phrases such as at least one. It is often shorter to subtract the probability of no successes from $1$.

Worked example 2: at least one

A biased coin has probability $0.3$ of landing heads on each toss. Tosses are independent. Find the probability of at least one head in three tosses.

The complement of “at least one head” is “no heads”, which means three tails. Since

$$\operatorname{P}(T)=1-0.3=0.7,$$

independence gives

$$\begin{aligned} \operatorname{P}(\text{at least one }H) &=1-\operatorname{P}(TTT)\\ &=1-(0.7)^3\\ &=1-0.343\\ &=\boxed{0.657}. \end{aligned}$$

Do not calculate $1-(0.3)^3$. Three heads is not the complement of at least one head.

Unions, intersections and Venn diagrams

For events $A$ and $B$:

• $A\cap B$ means $A$ and $B$, so both events occur;
• $A\cup B$ means $A$ or $B$, including the possibility that both occur;
• $A'$ means not $A$.

In probability, “or” is normally inclusive. The event $A\cup B$ contains the $A$ only region, the overlap, and the $B$ only region.

Adding $\operatorname{P}(A)$ and $\operatorname{P}(B)$ counts the overlap twice. Subtract it once to obtain the addition rule:

$$\boxed{\operatorname{P}(A\cup B) =\operatorname{P}(A)+\operatorname{P}(B)-\operatorname{P}(A\cap B)}.$$

Worked example 3: use the addition rule

In a group of students, $52\%$ study French, $38\%$ study Spanish and $21\%$ study both. One student is selected at random. Find the probability that the student:

1. studies French or Spanish;
2. studies neither language;
3. studies French but not Spanish.

Let $F$ and $S$ denote the two events.

First,

$$\begin{aligned} \operatorname{P}(F\cup S) &=\operatorname{P}(F)+\operatorname{P}(S)-\operatorname{P}(F\cap S)\\ &=0.52+0.38-0.21\\ &=\boxed{0.69}. \end{aligned}$$

Neither is the complement of the union:

$$\operatorname{P}((F\cup S)')=1-0.69=\boxed{0.31}.$$

French but not Spanish is the part of $F$ outside the overlap:

$$\operatorname{P}(F\cap S')=0.52-0.21=\boxed{0.31}.$$

Fill a Venn diagram from the centre

When an intersection is given, place it in the overlap first. Then subtract it from each event total to find the two outer regions. Find the region outside both last.

Worked example 4: find an unknown intersection

Suppose

$$\operatorname{P}(A)=0.63, \qquad \operatorname{P}(B)=0.48, \qquad \operatorname{P}(A\cup B)=0.79.$$

Find $\operatorname{P}(A\cap B)$.

Rearrange the addition rule:

$$\begin{aligned} \operatorname{P}(A\cap B) &=\operatorname{P}(A)+\operatorname{P}(B)-\operatorname{P}(A\cup B)\\ &=0.63+0.48-0.79\\ &=\boxed{0.32}. \end{aligned}$$

The answer is plausible because an intersection cannot exceed either event probability.

Self check 2

Given $\operatorname{P}(A)=0.7$, $\operatorname{P}(B)=0.4$ and $\operatorname{P}(A\cap B)=0.25$, find:

1. $\operatorname{P}(A\cup B)$;
2. $\operatorname{P}(A'\cap B)$;
3. $\operatorname{P}(A'\cap B')$.

Answer

$$\operatorname{P}(A\cup B)=0.7+0.4-0.25=\boxed{0.85}.$$

The part of $B$ outside $A$ is

$$\operatorname{P}(A'\cap B)=0.4-0.25=\boxed{0.15}.$$

Neither event occurs in the complement of the union, so

$$\operatorname{P}(A'\cap B')=1-0.85=\boxed{0.15}.$$

Mutually exclusive events

Events are mutually exclusive if they cannot occur together. Their intersection is empty:

$$\boxed{A\cap B=\varnothing} \qquad\Longleftrightarrow\qquad \boxed{\operatorname{P}(A\cap B)=0}.$$

The addition rule then simplifies to

$$\boxed{\operatorname{P}(A\cup B)=\operatorname{P}(A)+\operatorname{P}(B)}.$$

On one die roll, “roll a $2$” and “roll a $5$” are mutually exclusive. By contrast, “roll an even number” and “roll a number greater than $3$” overlap at $4$ and $6$, so they are not mutually exclusive.

Mutually exclusive does not mean independent

Mutually exclusive events cannot happen together. Independent events do not affect each other’s probabilities. If $A$ and $B$ have positive probability and are mutually exclusive, observing $A$ makes $B$ impossible, so they are dependent.

Tree diagrams

A tree diagram represents a sequence of events. Use three rules:

1. probabilities on branches leaving the same point sum to $1$;
2. multiply probabilities along a route;
3. add the probabilities of distinct routes that meet the requirement.

The multiplication rule in full is

$$\operatorname{P}(A\cap B) =\operatorname{P}(A)\operatorname{P}(B\mid A).$$

The symbol $\operatorname{P}(B\mid A)$ means the probability of $B$ given that $A$ has occurred. For independent events it equals $\operatorname{P}(B)$. Dependent trees are developed fully in conditional probability.

Worked example 5: with replacement

A bag contains $3$ red and $2$ blue counters. A counter is selected, replaced, and then a second counter is selected. Find the probability of obtaining one counter of each colour.

Replacement restores the original contents, so each selection has

$$\operatorname{P}(R)=\frac35, \qquad \operatorname{P}(B)=\frac25.$$

There are two mutually exclusive routes: $RB$ and $BR$.

$$\begin{aligned} \operatorname{P}(\text{one of each}) &=\operatorname{P}(RB)+\operatorname{P}(BR)\\ &=\frac35\times\frac25+\frac25\times\frac35\\ &=\frac6{25}+\frac6{25}\\ &=\boxed{\frac{12}{25}}. \end{aligned}$$

Multiplying gives one complete route. Adding combines the two alternative routes.

Worked example 6: without replacement

The same bag contains $3$ red and $2$ blue counters, but counters are not replaced. Find the probability that both selected counters have the same colour.

After the first selection, only $4$ counters remain. The required routes are $RR$ and $BB$:

$$\begin{aligned} \operatorname{P}(\text{same colour}) &=\frac35\times\frac24+\frac25\times\frac14\\ &=\frac3{10}+\frac1{10}\\ &=\boxed{\frac25}. \end{aligned}$$

The second branch probabilities change because the first counter is not replaced. This is dependence.

Self check 3

A box contains $4$ green and $6$ yellow balls. Two balls are selected without replacement. Find the probability of at least one green ball.

Answer

Use the complement, which is two yellow balls:

$$\begin{aligned} \operatorname{P}(\text{at least one green}) &=1-\operatorname{P}(YY)\\ &=1-\frac6{10}\times\frac59\\ &=1-\frac13\\ &=\boxed{\frac23}. \end{aligned}$$

Independent events

Events $A$ and $B$ are independent when knowing that one occurred does not change the probability of the other. The most useful test is

$$\boxed{\operatorname{P}(A\cap B)=\operatorname{P}(A)\operatorname{P}(B)}.$$

Do not assume independence merely because events sound unrelated. Use the information given or test the equation.

Worked example 7: test for independence

Suppose

$$\operatorname{P}(A)=0.4, \qquad \operatorname{P}(B)=0.7, \qquad \operatorname{P}(A\cup B)=0.82.$$

Determine whether $A$ and $B$ are independent.

First find the intersection:

$$\operatorname{P}(A\cap B)=0.4+0.7-0.82=0.28.$$

Now compare it with the product:

$$\operatorname{P}(A)\operatorname{P}(B)=(0.4)(0.7)=0.28.$$

The values are equal, so $A$ and $B$ are \boxed{\text{independent}}.

Worked example 8: find a probability using independence

Events $C$ and $D$ are independent, with

$$\operatorname{P}(C)=0.6 \qquad\text{and}\qquad \operatorname{P}(C\cup D)=0.8.$$

Find $\operatorname{P}(D)$.

Let $\operatorname{P}(D)=p$. Independence gives $\operatorname{P}(C\cap D)=0.6p$. Substitute into the addition rule:

$$0.8=0.6+p-0.6p.$$

Therefore

$$0.2=0.4p \qquad\Longrightarrow\qquad \boxed{p=0.5}.$$

This example requires both the addition rule and the independence condition. Adding $0.6+p$ alone would wrongly assume the events were mutually exclusive.

Self check 4

Events $A$ and $B$ satisfy $\operatorname{P}(A)=0.3$, $\operatorname{P}(B)=0.5$ and $\operatorname{P}(A\cap B)=0.15$.

1. Are $A$ and $B$ independent?
2. Are $A$ and $B$ mutually exclusive?

Answer

Since

$$\operatorname{P}(A)\operatorname{P}(B)=(0.3)(0.5)=0.15 =\operatorname{P}(A\cap B),$$

the events are independent.

They are not mutually exclusive because

$$\operatorname{P}(A\cap B)=0.15\ne0.$$

Systematic counting

When outcomes are equally likely, probability is often a counting problem. The multiplication principle says that if one stage has $m$ choices and a second stage has $n$ choices for each first choice, there are $mn$ possible ordered outcomes.

If order does not matter, combinations may be more efficient:

$$\binom nr=\frac{n!}{r!(n-r)!}.$$

Worked example 9: combinations in probability

A committee of $3$ is chosen at random from $5$ women and $4$ men. Find the probability that exactly $2$ women are chosen.

Every committee of $3$ is equally likely. The total number is

$$\binom93=84.$$

For exactly $2$ women, choose $2$ of the $5$ women and $1$ of the $4$ men:

$$\binom52\binom41=10\times4=40.$$

Hence

$$\operatorname{P}(\text{exactly }2\text{ women}) =\frac{40}{84}=\boxed{\frac{10}{21}}.$$

The choices are multiplied because both selections are required. There is no factor of $3!$ because a committee has no order.

Modelling and checking answers

A probability calculation is only as sound as its model. Before calculating, ask:

• Are the stated outcomes equally likely?
• Is sampling with replacement or without replacement?
• Is independence stated, justified by the mechanism, or testable from data?
• Does “or” include the overlap?
• Are the events exhaustive, so their probabilities should sum to $1$?

After calculating, check that the answer lies between $0$ and $1$. Also use bounds. For example,

$$0\leq \operatorname{P}(A\cap B)\leq \min\{\operatorname{P}(A),\operatorname{P}(B)\}.$$

If $\operatorname{P}(A)=0.2$, an answer of $0.35$ for $\operatorname{P}(A\cap B)$ is impossible because the overlap is contained within $A$.

Common misconceptions

Misconception	Correction
”Or” means add	Subtract the overlap unless the events are mutually exclusive.
”And” always means multiply the marginal probabilities	$\operatorname{P}(A)\operatorname{P}(B)$ is valid only for independent events. In general use a conditional branch probability.
Mutually exclusive means independent	They describe different ideas. Positive probability events cannot be both.
At least one means calculate every successful route	The complement of none is often shorter.
Without replacement leaves probabilities unchanged	The total and possibly the favourable count change after each selection.
A decimal from a calculator is automatically exact	Keep fractions or full precision during working, then round only at the end.

Mixed self check

In a group of $100$ people, $58$ read newspaper $A$, $45$ read newspaper $B$, and $18$ read neither. One person is selected at random.

1. Find the probability that the person reads at least one newspaper.
2. Find the probability that the person reads both newspapers.
3. Determine whether reading $A$ and reading $B$ are independent.
4. Find the probability that the person reads exactly one of the newspapers.

Answer

The probability of at least one is the complement of neither:

$$\operatorname{P}(A\cup B)=1-0.18=\boxed{0.82}.$$

Use the addition rule to find the overlap:

$$\begin{aligned} \operatorname{P}(A\cap B) &=0.58+0.45-0.82\\ &=\boxed{0.21}. \end{aligned}$$

For independence, compare:

$$\operatorname{P}(A)\operatorname{P}(B)=(0.58)(0.45)=0.261.$$

Since $0.21\ne0.261$, the events are not independent.

Exactly one consists of the two non-overlapping outer regions:

$$\begin{aligned} \operatorname{P}(\text{exactly one}) &=(0.58-0.21)+(0.45-0.21)\\ &=0.37+0.24\\ &=\boxed{0.61}. \end{aligned}$$

Key results

$$\begin{aligned} \operatorname{P}(A')&=1-\operatorname{P}(A),\\ \operatorname{P}(A\cup B)&=\operatorname{P}(A)+\operatorname{P}(B)-\operatorname{P}(A\cap B),\\ A,B\text{ mutually exclusive}&\Longrightarrow \operatorname{P}(A\cap B)=0,\\ A,B\text{ independent}&\Longleftrightarrow \operatorname{P}(A\cap B)=\operatorname{P}(A)\operatorname{P}(B). \end{aligned}$$

On a tree, multiply along routes and add alternative routes. Always decide whether later branches change before copying probabilities across the tree.

Next steps

Continue to conditional probability to formalise changing sample spaces, dependent events and reverse probability problems. Then study discrete random variables and the binomial distribution to turn repeated random processes into probability distributions. For the assumptions behind real applications, see probability modelling.