Probability foundations: rules, diagrams and conditional probability

Probability measures uncertainty on a scale from impossible to certain:

$$0\leq P(A)\leq 1.$$

Here $A$ is an event, such as obtaining an even number when a die is rolled. A probability may be written as a fraction, decimal or percentage, but exact fractions are usually best during a calculation.

This lesson secures the probability ideas assumed throughout A-level Mathematics. The central challenge is not arithmetic. It is translating words into the correct events and deciding whether to add, multiply or condition.

Prerequisites

You should be able to:

• simplify fractions;
• convert between fractions, decimals and percentages;
• use ratios and proportions;
• count systematically;
• read two-way tables.

Review exact arithmetic and ratio and proportion if needed.

Events and notation

The sample space, usually written $\Omega$, is the set of all possible outcomes. An event is a subset of that sample space.

For one fair six-sided die,

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

If $A$ is the event “the score is even”, then

$$A=\{2,4,6\},\qquad P(A)=\frac36=\frac12.$$

For equally likely outcomes,

$$P(A)=\frac{\text{number of outcomes in }A}{\text{total number of outcomes}}.$$

This formula is valid only when the outcomes being counted are equally likely. The outcomes “wins” and “loses” are two possibilities, but that does not make each probability $\frac12$.

Essential notation

Notation	Meaning
$P(A)$	probability that $A$ occurs
$A'$	complement of $A$: $A$ does not occur
$A\cap B$	both $A$ and $B$ occur
$A\cup B$	$A$ or $B$ or both occur
$P(A\mid B)$	probability of $A$, given that $B$ has occurred

In probability, or is normally inclusive. Thus $A\cup B$ includes the overlap $A\cap B$.

Complements and exhaustive events

An event and its complement are mutually exclusive and together cover the whole sample space. Therefore

$$P(A')=1-P(A).$$

Worked example 1: use a complement

The probability that a train is late is $0.18$. Find the probability that it is not late.

Let $L$ be the event that the train is late. Then

$$P(L')=1-P(L)=1-0.18=\boxed{0.82}.$$

Worked example 2: “at least one”

A fair coin is tossed three times. Find the probability of at least one head.

The direct event contains several cases: exactly one, exactly two or exactly three heads. Its complement is simpler:

$$\text{no heads}=\text{three tails}.$$

The tosses are independent, so

$$P(\text{three tails})=\left(\frac12\right)^3=\frac18.$$

Hence

$$P(\text{at least one head})=1-\frac18=\boxed{\frac78}.$$

The phrase at least one often signals a useful complement, but always check whether the complementary event really is easier.

Experimental probability

If an event occurs $r$ times in $n$ trials, its relative frequency is

$$\frac rn.$$

This estimates the underlying probability. It is not guaranteed to equal it. Over many repeated trials, relative frequency tends to become more stable around the modelled probability.

Worked example 3: estimate and predict

A drawing pin lands point-up $93$ times in $150$ trials.

An estimate for the probability of point-up is

$$\hat p=\frac{93}{150}=0.62.$$

Using this estimate, the expected number of point-up results in another $400$ trials is

$$400(0.62)=\boxed{248}.$$

This is a modelled expectation, not a promise of exactly $248$.

Misconception: short-run balancing

If a fair coin gives five heads in succession, a tail is not “due”. Provided the tosses are independent,

$$P(\text{tail on the next toss})=\frac12.$$

Long-run stability does not force results to compensate immediately.

Constructing sample spaces

A complete, non-repeating sample space prevents missed or double-counted cases. Use an ordered list, table or tree.

Worked example 4: a sample-space table

Two fair four-sided spinners are numbered $1,2,3,4$. Find the probability that their total is $5$.

There are $4\times4=16$ equally likely ordered pairs. The favourable pairs are

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

Therefore

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

The totals $2,3,\ldots,8$ are not equally likely. For example, only $(1,1)$ gives $2$, while four pairs give $5$.

Worked example 5: counting without replacement

A box contains four cards labelled $A,B,C,D$. Two different cards are selected in order without replacement. Find the probability that one is $A$.

There are $4\times3=12$ equally likely ordered outcomes. Those containing $A$ are

$$AB,AC,AD,BA,CA,DA.$$

Thus

$$P(\text{one is }A)=\frac6{12}=\boxed{\frac12}.$$

Here “one is $A$” means exactly one, because $A$ cannot be selected twice without replacement.

Self-check 1

1. A fair die is rolled. Find $P(\text{score greater than }4)$.
2. Two fair coins are tossed. Find $P(\text{exactly one head})$.
3. The probability of rain is $0.35$. Find the probability of no rain.

Answers

1. The favourable scores are $5$ and $6$, so $P=\frac26=\frac13$.
2. The equally likely outcomes are $HH,HT,TH,TT$. Two give exactly one head, so $P=\frac24=\frac12$.
3. $1-0.35=0.65$.

The addition rule

Adding $P(A)$ and $P(B)$ counts outcomes in the overlap twice. Subtract the overlap once:

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

Events are mutually exclusive if they cannot occur together. Then $A\cap B$ is impossible, so

$$P(A\cup B)=P(A)+P(B).$$

Worked example 6: mutually exclusive events

A fair die is rolled. Let $A$ be “score $1$” and $B$ be “score greater than $4$”. These events cannot happen together, so

$$P(A\cup B)=\frac16+\frac26=\boxed{\frac12}.$$

Worked example 7: overlapping events

One card is selected from a standard pack of $52$. Find the probability that it is a heart or a king.

There are $13$ hearts and $4$ kings. The king of hearts belongs to both sets, so

$$P(H\cup K)=\frac{13}{52}+\frac4{52}-\frac1{52} =\frac{16}{52}=\boxed{\frac4{13}}.$$

Simply calculating $(13+4)/52$ counts the king of hearts twice.

Venn diagrams

A Venn diagram represents events as regions. Fill a diagram from the overlap outwards:

1. enter $A\cap B$;
2. calculate the parts belonging only to $A$ and only to $B$;
3. calculate the region outside both events;
4. check that all regions total the size of the universal set.

Worked example 8: recover every region

In a group of $80$ students, $46$ study French, $35$ study Spanish and $18$ study both. One student is selected at random.

The four disjoint regions are:

$$\begin{aligned} \text{both}&=18,\\ \text{French only}&=46-18=28,\\ \text{Spanish only}&=35-18=17,\\ \text{neither}&=80-(18+28+17)=17. \end{aligned}$$

Therefore

$$P(F\cup S)=\frac{18+28+17}{80}=\boxed{\frac{63}{80}}$$

and

$$P(F'\cap S)=P(\text{Spanish only})=\boxed{\frac{17}{80}}.$$

Notice that $F'\cap S$ means outside $F$ but inside $S$. A prime changes the region, not merely the notation.

The multiplication rule and conditional probability

For events $A$ and $B$,

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

After $A$ has occurred, $P(B\mid A)$ uses a reduced sample space containing only outcomes in $A$.

Rearranging gives the conditional probability formula

$$\boxed{P(B\mid A)=\frac{P(A\cap B)}{P(A)}},\qquad P(A)>0.$$

Worked example 9: condition on a subgroup

In the group from Worked example 8, find the probability that a student studies Spanish, given that the student studies French.

The condition restricts attention to the $46$ French students. Of those, $18$ also study Spanish. Therefore

$$P(S\mid F)=\frac{18}{46}=\boxed{\frac9{23}}.$$

The denominator is $46$, not $80$, because the phrase “given that” changes the sample space.

Worked example 10: multiply along a route

A bag contains $5$ red and $3$ blue counters. Two counters are taken without replacement. Find the probability that both are red.

The first counter is red with probability $5/8$. Given that it was red, $4$ red counters remain among $7$ counters. Hence

$$P(R_1\cap R_2)=P(R_1)P(R_2\mid R_1) =\frac58\times\frac47 =\boxed{\frac5{14}}.$$

The second fraction changes because there is no replacement.

Independence

Events $A$ and $B$ are independent if knowing that one occurred does not change the probability of the other:

$$P(B\mid A)=P(B).$$

Equivalently,

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

This product is a test for independence, not a rule to apply automatically.

Worked example 11: test independence

Suppose

$$P(A)=0.4,\qquad P(B)=0.5,\qquad P(A\cap B)=0.2.$$

Since

$$P(A)P(B)=0.4(0.5)=0.2=P(A\cap B),$$

$A$ and $B$ are independent.

If instead $P(A\cap B)=0.15$, they would not be independent.

Mutually exclusive is not independent

If two events with non-zero probabilities are mutually exclusive, occurrence of one makes the other impossible. Therefore they are dependent.

For example, on one die roll, “odd” and “even” are mutually exclusive. But

$$P(\text{odd}\cap\text{even})=0 \ne \frac12\times\frac12.$$

Mutually exclusive concerns whether events can occur together. Independent concerns whether one changes the probability of the other.

Probability trees

A tree diagram displays successive events. Use two rules:

• probabilities on branches leaving the same node sum to $1$;
• multiply probabilities along a route, then add probabilities of distinct routes that satisfy the event.

Worked example 12: independent trials

A biased coin has $P(H)=0.6$ and is tossed twice independently. Find the probability of exactly one head.

There are two suitable routes:

$$HT\quad\text{or}\quad TH.$$

Their probabilities are

$$P(HT)=0.6(0.4)=0.24, \qquad P(TH)=0.4(0.6)=0.24.$$

The routes are mutually exclusive, so add:

$$P(\text{exactly one head})=0.24+0.24=\boxed{0.48}.$$

Worked example 13: without replacement

A bag contains $4$ green and $2$ yellow counters. Two are selected without replacement. Find the probability that they have different colours.

There are two routes:

$$GY\quad\text{or}\quad YG.$$

After one counter is removed, $5$ remain. Thus

$$\begin{aligned} P(GY)&=\frac46\times\frac25=\frac4{15},\\ P(YG)&=\frac26\times\frac45=\frac4{15}. \end{aligned}$$

Therefore

$$P(\text{different colours})=\frac4{15}+\frac4{15} =\boxed{\frac8{15}}.$$

As a check, the complementary event is two counters of the same colour:

$$\frac46\times\frac35+\frac26\times\frac15 =\frac6{15}+\frac1{15}=\frac7{15},$$

and $8/15+7/15=1$.

Worked example 14: reverse a tree using algebra

A factory uses machines $A$ and $B$. Machine $A$ makes $60\%$ of the items and $2\%$ of its items are defective. Machine $B$ makes the rest and $5\%$ of its items are defective. Given that an item is defective, find the probability it came from $B$.

Let $D$ mean defective. First find the probabilities of the two defective routes:

$$P(A\cap D)=0.60(0.02)=0.012,$$ $$P(B\cap D)=0.40(0.05)=0.020.$$

Hence

$$P(D)=0.012+0.020=0.032.$$

Restricting attention to defective items gives

$$P(B\mid D)=\frac{P(B\cap D)}{P(D)} =\frac{0.020}{0.032} =\boxed{0.625}.$$

Although $B$ makes fewer items, its higher defect rate means it is responsible for most defective items.

Two-way tables

Two-way tables are especially useful when data are given as frequencies. Add row and column totals before calculating probabilities. For a conditional probability, use the total of the conditioned row or column as the denominator.

Worked example 15: table and condition

The table records whether $120$ people cycle to work.

	Cycles	Does not cycle	Total
Under 40	$32$	$28$	$60$
40 or over	$18$	$42$	$60$
Total	$50$	$70$	$120$

For a randomly selected person,

$$P(\text{under 40 and cycles})=\frac{32}{120}=\frac4{15}.$$

Given that the person cycles, the relevant column contains $50$ people, so

$$P(\text{under 40}\mid\text{cycles})=\frac{32}{50}=\boxed{\frac{16}{25}}.$$

By contrast,

$$P(\text{cycles}\mid\text{under 40})=\frac{32}{60}=\frac8{15}.$$

Reversing the condition usually changes the probability.

Choosing the correct operation

Translate the event before calculating.

Wording or structure	Typical operation	Question to ask
not $A$	complement	Is $1-P(A)$ simpler?
$A$ or $B$	addition rule	Is there an overlap to subtract?
$A$ and then $B$	multiplication rule	Does the second probability depend on the first?
given $B$	conditional probability	What is the reduced sample space?
at least one	often a complement	Is “none” a single simple case?

Words alone do not determine an operation. In “red or blue” the events may be mutually exclusive for one counter. In “studies French or Spanish” they may overlap.

Common misconceptions

Assuming outcomes are equally likely

Count outcomes only after establishing equal likelihood. A spinner’s sectors are equally likely only if they have equal angles and the physical model supports that assumption.

Adding for every “or”

Use

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

Omit the overlap only when the events are mutually exclusive.

Multiplying unchanged probabilities without justification

With replacement, probabilities usually stay constant. Without replacement, they usually change. Independence must be stated, implied by a valid model or verified.

Confusing $P(A\mid B)$ with $P(B\mid A)$

The event after the vertical bar is the information already known. It determines the denominator.

Rounding too early

Keep fractions or full calculator values until the end. Early rounding can noticeably distort probabilities built from several branches.

Mixed self-check

1. Given $P(A)=0.7$, $P(B)=0.4$ and $P(A\cap B)=0.25$, find $P(A\cup B)$.
2. A fair die is rolled twice. Find the probability of at least one $6$.
3. A bag contains $3$ red and $7$ blue counters. Two are selected without replacement. Find the probability that both are blue.
4. In a class, $12$ of the $20$ girls and $9$ of the $15$ boys study physics. A physics student is chosen at random. Find the probability that the student is a girl.
5. Events $C$ and $D$ satisfy $P(C)=0.3$, $P(D)=0.6$ and $P(C\cap D)=0.18$. Are they independent?
6. Explain why two mutually exclusive events with positive probabilities cannot be independent.

Answers

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

2. The complement is no sixes: $1-\left(\frac56\right)^2=1-\frac{25}{36}=\boxed{\frac{11}{36}}.$

3. $\frac7{10}\times\frac69=\boxed{\frac7{15}}.$

4. There are $12+9=21$ physics students, of whom $12$ are girls: $P(\text{girl}\mid\text{physics})=\frac{12}{21}=\boxed{\frac47}.$

5. Yes, because $P(C)P(D)=0.3(0.6)=0.18=P(C\cap D).$

6. If $C$ and $D$ are mutually exclusive, $P(C\cap D)=0$. If both probabilities are positive, $P(C)P(D)>0$, so $P(C\cap D)\ne P(C)P(D)$ and the independence condition fails.

Final checklist

You should now be able to:

• describe sample spaces and events using set notation;
• calculate theoretical and experimental probabilities;
• use complements, including for “at least one” questions;
• apply the general addition rule and recognise mutually exclusive events;
• apply the multiplication rule and recognise independent events;
• calculate conditional probabilities from diagrams and tables;
• build and interpret probability trees with and without replacement;
• distinguish independence from mutual exclusivity;
• check that every probability lies between $0$ and $1$.

Continue to probability at A-level and conditional probability. You will also use these foundations in choosing a probability distribution and checking answers.