M1StatisticsFoundation

Probability Basics

Q: Why subtract the intersection in the addition law?

Because outcomes in both events are included in P(A) and again in P(B). Subtracting one copy leaves each outcome counted exactly once.

Q: What does "at least one" translate to?

It is the complement of "none", so P( at least one) = 1 - P( none). That is almost always faster than adding the separate cases.

Probability measures how likely an outcome is, on a scale from 0 (impossible) to 1 (certain). Everything in this chapter follows from counting outcomes carefully — and from one law that stops you counting the same outcome twice.

30 min Probability

The big picture

The single most useful habit here is drawing a Venn diagram and filling in the $\textbf{intersection first}$ . Almost every mistake in probability comes from double-counting or from misreading which region a description refers to, and a filled diagram makes both visible. It also pays off well beyond this lesson: conditional probability is reading a Venn diagram with the sample space restricted, and independence is a statement about how two regions overlap. Getting fluent with the diagram now means the harder ideas that follow are mostly translation rather than new mathematics.

What you'll be able to do

Use sample spaces and equally likely outcomes to find probabilities
Apply the addition law, including for mutually exclusive events
Construct and read Venn diagrams for two and three events
Use the complement to simplify "at least one" problems

Sample spaces and equally likely outcomes

The $\textbf{sample space}$ is the set of all possible outcomes. When outcomes are equally likely, the probability of an event is the number of outcomes in it divided by the total.

Listing the sample space systematically is what prevents omissions. For two dice, a $6\times6$ grid of the 36 ordered pairs makes every event easy to count, and it makes clear that $\left(2,5\right)$ and $\left(5,2\right)$ are different outcomes.

The equally-likely assumption has to be earned. It holds for a fair die but not for the $\textit{totals}$ of two dice, where 7 arises six ways and 12 only one.

P(A) = \dfrac{n(A)}{n(S)}

Equally likely outcomes only.

0 \leq P(A) \leq 1

The probability scale.

Example 1Two fair dice are thrown. Find the probability the total is 9.

1Sample space has

6\times6 = 36

equally likely ordered pairs.

2Totals of 9:

\left(3,6\right), \left(4,5\right), \left(5,4\right), \left(6,3\right)

— four ways.

P = \tfrac{4}{36}

Answer

\tfrac{1}{9}

Tip — Count ordered outcomes. Treating $\left(3,6\right)$ and $\left(6,3\right)$ as one outcome halves several probabilities and is a standard error.

The addition law

For any two events, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . The subtraction is the whole content of the law: outcomes in both events get counted once in $P(A)$ and again in $P(B)$ , so one copy must be removed.

Events are $\textbf{mutually exclusive}$ when they cannot both happen, so $P(A \cap B) = 0$ and the law simplifies to $P(A) + P(B)$ . On a Venn diagram their circles do not overlap.

The $\textbf{complement}$ $A'$ is everything not in $A$ , and $P(A') = 1 - P(A)$ . It is the shortcut for any "at least one" question, since the opposite of "at least one" is "none".

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

The addition law.

P(A \cup B) = P(A)+P(B) \text{ if exclusive}

No overlap to remove.

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

The complement.

Example 1In a class,

P(\text{plays football}) = 0.6

P(\text{plays tennis}) = 0.3

and

P(\text{both}) = 0.2

. Find the probability a student plays at least one.

1Apply the addition law:

0.6 + 0.3 - 0.2

2The 0.2 is removed because those students appear in both of the first two figures.

Answer

0.7

Example 2Using the same figures, find the probability a student plays neither.

1"Neither" is the complement of "at least one".

1 - 0.7

Answer

0.3

Why this works

Without the subtraction the answer would be $0.9$ , which counts the 20% who play both twice. Whenever an "or" probability comes out suspiciously high, check whether the overlap was removed.

Venn diagrams

A Venn diagram shows events as overlapping regions inside a rectangle representing the sample space. The reliable method is to fill in the $\textbf{intersection first}$ , then work outwards subtracting what is already placed.

Filling outwards prevents the most common error: writing the total for $A$ inside the part of $A$ that excludes $B$ . The region " $A$ only" holds $n(A) - n(A \cap B)$ , not $n(A)$ .

Once complete, every probability is read off by counting regions, and the language maps directly: " $A$ and $B$ " is the overlap, " $A$ or $B$ " is everything in either circle, "neither" is outside both.

Example 1Of 50 students, 28 study French, 22 study German and 10 study both. How many study neither?

1Intersection first: 10 in the overlap.

2French only:

28 - 10 = 18

. German only:

22 - 10 = 12

3Total studying at least one:

18 + 10 + 12 = 40

4Neither:

50 - 40

Answer10 students study neither.

Tip — Check that all your regions sum to the total. If they do not, a value has been placed in the wrong region or an overlap has been double-counted.

Three events, and useful language

Three-set Venn diagrams have eight regions and the same rule applies with more care: fill the triple intersection first, then the pairwise overlaps (subtracting the centre), then the singles, then outside.

Some phrases are worth translating explicitly. "Exactly one" means the three single regions added. "At least two" means the three pairwise regions plus the centre. "At most one" is the complement of "at least two".

Reading the phrase into regions before calculating is faster and safer than trying to build a formula for each case.

Example 1In a survey,

n(A \cap B \cap C) = 5

n(A \cap B) = 12

n(A \cap C) = 9

n(B \cap C) = 8

. Find the number in exactly two of the three sets.

1Each pairwise count includes the 5 in the centre.

A \cap B

only:

12 - 5 = 7

A \cap C

only:

9 - 5 = 4

B \cap C

only:

8 - 5 = 3

3Sum the three:

7 + 4 + 3

Answer14 are in exactly two sets.

Why this works

Notice that $n(A \cap B) = 12$ means "in both $A$ and $B$ ", which $\textit{includes}$ those also in $C$ . "Exactly $A$ and $B$ " is a different, smaller region — questions distinguish the two deliberately.

Think like an examiner

Draw the Venn diagram even when not asked. It is quick and prevents the double-counting errors that cost most marks here.
Fill the intersection first and work outwards, checking the regions sum to the total.
Use the complement for "at least one" rather than adding many cases.
Give probabilities as exact fractions or decimals as the question asks, and never above 1.

Common misconceptions

Writing

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

when the events overlap.

Subtract the intersection, or outcomes in both are counted twice.

Placing

n(A)

in the "

A

only" region.

A

only" is

n(A) - n(A\cap B)

. The total for

A

spans both regions.

Treating

\left(2,5\right)

and

\left(5,2\right)

as the same dice outcome.

They are distinct ordered outcomes in the 36-cell sample space.

Reading

n(A\cap B)

as "in

A

and

B

but not

C

It includes the triple intersection. Subtract the centre to get "exactly

A

and

B

Probability laws

P(A) = \tfrac{n(A)}{n(S)}

Equally likely outcomes.

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

Addition law.

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

Complement.

P(A\cap B) = 0

Mutually exclusive.

Stretch yourself

In a group of 120 people, 64 read newspaper $A$ , 50 read $B$ , 42 read $C$ , 22 read both $A$ and $B$ , 18 read both $A$ and $C$ , 15 read both $B$ and $C$ , and 8 read all three. How many read none of the three?

Hint — Fill the Venn diagram from the centre outwards, remembering each pairwise figure includes the centre.

Reveal solution

SolutionCentre: 8. Exactly

A\cap B

22-8 = 14

. Exactly

A\cap C

18-8 = 10

. Exactly

B\cap C

15-8 = 7

. Now the singles:

A

only

= 64 - 14 - 10 - 8 = 32

;

B

only

= 50 - 14 - 7 - 8 = 21

;

C

only

= 42 - 10 - 7 - 8 = 17

. Total reading at least one

= 32+21+17+14+10+7+8 = 109

. So

120 - 109 = 11

read none. The same result follows from inclusion–exclusion:

64+50+42-22-18-15+8 = 109

, which is worth using as a check — the pattern of adding singles, subtracting pairs and adding back the triple is exactly what the diagram is doing geometrically.

\blacksquare

Questions students ask

Why subtract the intersection in the addition law?

Because outcomes in both events are included in

P(A)

and again in

P(B)

. Subtracting one copy leaves each outcome counted exactly once.

Are mutually exclusive and independent the same?

No, and they are close to opposites. Mutually exclusive means they cannot both happen; independent means one happening does not change the probability of the other. Two events with non-zero probability cannot be both.

When can I use

\tfrac{n(A)}{n(S)}

Only when all outcomes in the sample space are equally likely. Dice totals are not equally likely even though individual faces are.

What does "at least one" translate to?

It is the complement of "none", so

P(\text{at least one}) = 1 - P(\text{none})

. That is almost always faster than adding the separate cases.

Key takeaways

Probability runs from 0 to 1, and equals favourable over total when outcomes are equally likely.
The addition law subtracts the intersection to avoid double counting.
Mutually exclusive events have zero intersection; the complement gives $1 - P(A)$ .
Fill Venn diagrams from the intersection outwards and check the regions total correctly.
Translate phrases like "exactly one" and "at least two" into regions before calculating.