NDA Maths · Probability

Event Algebra & the Addition Rule

Combining events with set operations — the addition rule for unions, complements for 'neither', and what mutually exclusive and exhaustive really mean.

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Why this matters

Once you can count outcomes, the next step is combining events: 'A or B', 'A and B', 'neither'. The addition rule P(A or B) = P(A) + P(B) - P(A and B) is the workhorse of this 21-question subtopic, and the same diagram answers 'neither', 'exactly one', and 'at least one'. Get the inclusion-exclusion subtraction and the mutually-exclusive vs independent distinction right and these are reliable marks.

Test yourself — a quick 13-question Probability recall quiz, instant score.

Concept 1 of 5: Events as sets: union, intersection, complement

Events are subsets of the sample space, so the language of sets applies directly.

A \cup B

is "A or B (or both)";

A \cap B

is "A and B together";

A'

is "A does not happen". A Venn diagram turns every two-event question into filling in regions.

Definition

For events $A, B \subseteq S$ :

Union $A \cup B$ — occurs when at least one of them occurs.
Intersection $A \cap B$ — occurs when both occur.
Complement $A'$ (or $\bar{A}$ ) — occurs when $A$ does not.
Exactly one of $A, B$ is $(A \cap B') \cup (A' \cap B)$ ; neither is $A' \cap B'$ .
De Morgan: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$ .

Diagram · union, intersection, complement

Union (A ∪ B) is everything in A or B; intersection (A ∩ B) is the lens in both; complement (A′) is everything outside A. These map "or / and / not" straight onto regions.

Worked example

In a group,

A

is the event "likes tea" and

B

is "likes coffee". Express in set notation: (i) likes both, (ii) likes at least one, (iii) likes neither.

Practice this concept4 quick reps

$A \cup B$ is INCLUSIVE "or" — it contains the overlap

"A or B" in probability always means "A, or B, or both". The exclusive "exactly one" is a different event,

(A \cap B') \cup (A' \cap B)

Read "and" as intersection, "or" as union — do not swap

A \cap B

is the smaller event (both must hold);

A \cup B

is the larger (one suffices). Mixing them up flips the whole calculation.

Drill 6 more on events as sets: union, intersection, complement

Concept 2 of 5: The addition rule (inclusion-exclusion)

To find

P(A \text{ or } B)

you add

P(A)

and

P(B)

— but the overlap

A \cap B

has now been counted twice, so subtract it once.

Definition

For any two events, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . For three events the pattern continues: $P(A \cup B \cup C) = \sum P(A) - \sum P(A \cap B) + P(A \cap B \cap C)$ — add singles, subtract pairs, add the triple.

Addition rule

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

$P(A \cap B)$ probability both occur — subtracted to undo double-counting

Visualization · two events in the sample space

P(A): 0.5P(B): 0.4P(A∩B): 0.2

P(A∪B) = 0.7neither = 0.3exactly one = 0.5

P(A∪B) = P(A) + P(B) − P(A∩B): the lens is counted once, not twice. "Neither" is everything outside both circles, 1 − P(A∪B). The overlap is held inside its feasible range, so it never claims more than the smaller event or less than the forced minimum.

Worked example

The probability a family owns a car is

0.6

, owns a bike is

0.5

, and owns both is

0.3

. What is the probability it owns at least one?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2025 · Q103Easy

Example 2 · Probability · Event Algebra — Inclusion-Exclusion, Mutually Exclusive, Exhaustive

The probability that a family owns a laptop is 0.68; that it also owns a desktop is 0.56. If the probability that it owns both is 0.48, then what is the probability that a randomly selected family owns a laptop or a desktop?

Do not forget to subtract $P(A \cap B)$

P(A) + P(B)

alone overcounts the overlap. The only time you may skip the subtraction is when

A

and

B

are mutually exclusive (overlap

= 0

Three events need the full inclusion-exclusion, not just three single terms

Subtract all three pairwise intersections, then add back the triple intersection. Stopping after the singles or after the pairs gives a wrong answer.

Drill 4 more on the addition rule (inclusion-exclusion)

Concept 3 of 5: "Neither" and the complement of a union

"Neither A nor B" is just the complement of "A or B". So compute the union, then subtract from 1. De Morgan's law makes this exact: the outside of

A \cup B

A' \cap B'

Definition

$P(\text{neither } A \text{ nor } B) = P(A' \cap B') = 1 - P(A \cup B)$ . Equivalently $1 - \big(P(A) + P(B) - P(A \cap B)\big)$ . This is the fastest route for percentage "how many do neither" word problems.

Complement of a union (De Morgan)

P(A' \cap B') = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A \cap B)

$A' \cap B'$ the "neither" region — outside both circles

Diagram · "neither" = (A ∪ B)′

Everything shaded lies outside both circles — that's "neither A nor B", the complement of the union: P(neither) = 1 − P(A ∪ B). By De Morgan's law it's the same as A′ ∩ B′.

Worked example

In a town, 70% of households take a newspaper and 50% take a magazine; 30% take both. What percentage take neither?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q19Easy

Example 3 · Probability · Event Algebra — Inclusion-Exclusion, Mutually Exclusive, Exhaustive

In a school, 50% students play cricket and 40% play football. If 10% of students play both the games, then what per cent of students play neither cricket nor football ?

"Neither" is $1 - P(A \cup B)$ , not $1 - P(A) - P(B)$

Skipping the

+P(A \cap B)

term double-subtracts the overlap and understates the answer. Compute the union correctly first, then take the complement.

De Morgan flips the operation: complement of a union is an intersection

(A \cup B)' = A' \cap B'

(neither), while

(A \cap B)' = A' \cup B'

(not both). Using the wrong one swaps two different events.

Drill 2 more on "neither" and the complement of a union

Concept 4 of 5: Mutually exclusive (disjoint) events

Two events are mutually exclusive when they cannot both happen — their intersection is empty. Then the overlap term vanishes and the addition rule simplifies to plain addition.

Definition

$A$ and $B$ are mutually exclusive (disjoint) if $A \cap B = \varnothing$ , so $P(A \cap B) = 0$ . Then $P(A \cup B) = P(A) + P(B)$ . Mutually exclusive is NOT the same as independent: if two events with positive probability are mutually exclusive, the occurrence of one rules out the other, so they are in fact dependent.

Addition rule for mutually exclusive events

A \cap B = \varnothing \;\Rightarrow\; P(A \cup B) = P(A) + P(B)

$\varnothing$ the impossible event — the two cannot co-occur

Worked example

Two mutually exclusive events have

P(A) = 0.3

and

P(B) = 0.5

. Find

P(A \cup B)

and

P(A' \cap B')

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q119Moderate

Example 4 · Probability · Event Algebra — Inclusion-Exclusion, Mutually Exclusive, Exhaustive

Let

A

and

B

are two mutually exclusive events with

P(A)=\dfrac{1}{3}

and

P(B)=\dfrac{1}{4}

. What is the value of

P(\bar{A}\cap\bar{B})

Mutually exclusive $\ne$ independent — opposite ideas

Mutually exclusive means

P(A \cap B) = 0

; independent means

P(A \cap B) = P(A)P(B)

. For events with positive probability these cannot hold at once. Questions often test exactly this confusion.

Only drop the overlap term when you are TOLD the events are mutually exclusive

If a problem does not state disjointness (or give

P(A \cap B) = 0

), you must keep the

-P(A \cap B)

term in the addition rule.

Drill 1 more on mutually exclusive (disjoint) events

Concept 5 of 5: Exhaustive events (and probabilities that sum to 1)

A set of events is exhaustive when together they cover the whole sample space — one of them must occur. If they are also mutually exclusive, their probabilities add up to exactly 1, which turns a ratio between them into a solvable equation.

Definition

Events $A_1, \dots, A_n$ are exhaustive if $A_1 \cup \dots \cup A_n = S$ . If they are also mutually exclusive (a partition of $S$ ), then $P(A_1) + \dots + P(A_n) = 1$ . Most PYQs give the probabilities as a ratio (e.g. $2P(A) = 3P(B) = 4P(C)$ ) and use the sum-to-1 condition to solve.

Mutually exclusive AND exhaustive

A_1 \cup \dots \cup A_n = S \;\text{ and disjoint} \;\Rightarrow\; \sum_{i} P(A_i) = 1

partitionmutually exclusive + exhaustive: exactly one event occurs

Diagram · exhaustive events tile the sample space

The three events leave no gap and no overlap — they exhaust S. When events are both exhaustive and mutually exclusive (a partition), their probabilities add to exactly 1: 0.5 + 0.3 + 0.2 = 1. This is the backbone of the total-probability rule.

Worked example

A, B, C

are mutually exclusive and exhaustive with

P(A) = 2P(B) = 3P(C)

. Find

P(A)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q105Moderate

Example 5 · Probability · Event Algebra — Inclusion-Exclusion, Mutually Exclusive, Exhaustive

A, B, C, D are mutually exclusive and exhaustive. If

2P(A)=3P(B)=4P(C)=5P(D)

, then what is

77P(A)

equal to?

Exhaustive alone does not give sum $= 1$

If the events overlap,

\sum P(A_i)

exceeds 1 by the overlaps. The probabilities add to exactly 1 only when the events are BOTH exhaustive AND mutually exclusive (a partition).

Turn a chained ratio into one variable before summing

2P(A) = 3P(B) = 4P(C) = k

means

P(A) = \tfrac{k}{2}

P(B) = \tfrac{k}{3}

P(C) = \tfrac{k}{4}

. Set the sum to 1 to find

k

; a common slip is reading the ratio as

P(A):P(B):P(C) = 2:3:4

(it is actually the reciprocals).

Drill 8 more on exhaustive events (and probabilities that sum to 1)

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The addition rule (inclusion-exclusion)
Addition rule
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
"Neither" and the complement of a union
Complement of a union (De Morgan)
$P(A' \cap B') = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A \cap B)$
Mutually exclusive (disjoint) events
Addition rule for mutually exclusive events
$A \cap B = \varnothing \;\Rightarrow\; P(A \cup B) = P(A) + P(B)$
Exhaustive events (and probabilities that sum to 1)
Mutually exclusive AND exhaustive
$A_1 \cup \dots \cup A_n = S \;\text{ and disjoint} \;\Rightarrow\; \sum_{i} P(A_i) = 1$

Watch out for (10)

$A \cup B$ is INCLUSIVE "or" — it contains the overlap
→ Events as sets: union, intersection, complement
Read "and" as intersection, "or" as union — do not swap
→ Events as sets: union, intersection, complement
Do not forget to subtract $P(A \cap B)$
→ The addition rule (inclusion-exclusion)
Three events need the full inclusion-exclusion, not just three single terms
→ The addition rule (inclusion-exclusion)
"Neither" is $1 - P(A \cup B)$ , not $1 - P(A) - P(B)$
→ "Neither" and the complement of a union
De Morgan flips the operation: complement of a union is an intersection
→ "Neither" and the complement of a union
Mutually exclusive $\ne$ independent — opposite ideas
→ Mutually exclusive (disjoint) events
Only drop the overlap term when you are TOLD the events are mutually exclusive
→ Mutually exclusive (disjoint) events
Exhaustive alone does not give sum $= 1$
→ Exhaustive events (and probabilities that sum to 1)
Turn a chained ratio into one variable before summing
→ Exhaustive events (and probabilities that sum to 1)

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