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Class 12 Statistics Notes · GSEB

Addition Law

Probability — learn the general addition law P(A∪B), mutually exclusive, exhaustive, and derived results with step-by-step demonstrations. GSEB Class 12 Statistics notes.

Last updated: 25 Aug 2026

Notes

Addition Law of Probability

The addition law tells us how to find P(A ∪ B) — the probability that at least one of two events occurs. The key insight: subtract the overlap to avoid double counting.

General Addition Law

General Addition Law

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Why subtract P(A ∩ B)?

Imagine counting all students who play cricket OR football. If you add cricket players + football players, those who play BOTH get counted twice. Subtracting the overlap corrects this.

Mutually Exclusive Addition Law

Mutually Exclusive
When A ∩ B = ∅ (no overlap), the addition law simplifies — just add the probabilities.

Mutually Exclusive Addition

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

Example: Mutually Exclusive

A coin toss: A = heads, B = tails. P(A ∪ B) = 1/2 + 1/2 = 1. They cannot both happen — no overlap to subtract.

Example: NOT Mutually Exclusive

Drawing a card: A = heart, B = king. P(A ∩ B) = 1/52 (king of hearts). Must subtract!

Mutually Exclusive AND Exhaustive

Mutually Exclusive + Exhaustive
When A ∩ B = ∅ AND A ∪ B = S, the two events partition the entire sample space. Every outcome belongs to exactly one of them.

Condition

A∩B=∅ and A∪B=SA \cap B = \emptyset \text{ and } A \cup B = S

Key Result

P(A)+P(B)=1P(A) + P(B) = 1

Die: A = {1,2,3}, B = {4,5,6}

A ∩ B = ∅ (no common outcomes) and A ∪ B = S. So P(A) + P(B) = 3/6 + 3/6 = 1.

Think about it

If you know P(A) for mutually exclusive and exhaustive events, you automatically know P(B) = 1 − P(A). No need to count B separately!

Explore: Step-by-Step Addition Law

Click "The Question" to see the problem, then walk through 5 steps. Toggle "Mutually Exclusive" to see how the formula simplifies.

Addition Law — Step-by-Step
SAB

The Question

A = {1,2,3,4}, B = {3,4,5,6} — find P(A ∪ B)

4/6
P(A)
4/6
P(B)
2/6
P(A∩B)

Derived Results

Neither Happens

Neither A nor B

P(A′∩B′)=P((A∪B)′)=1−P(A∪B)P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)

Probability that neither A nor B occurs = complement of at least one.

At Least One is True

Not Both

P(A′∪B′)=P((A∩B)′)=1−P(A∩B)P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)

Probability that at least one of A' or B' is true = complement of both.

Only One Event Occurs

Exactly One

P((A−B)∪(B−A))=P(A)+P(B)−2P(A∩B)P((A-B) \cup (B-A)) = P(A) + P(B) - 2P(A \cap B)

Outcomes in A or B but NOT both. Also = P(A∪B) − P(A∩B).

Three Events (General)

Include-Exclude

P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)P(A \cup B \cup C) = P(A)+P(B)+P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)

For 3 mutually exclusive events: P(A∪B∪C) = P(A) + P(B) + P(C).

Solved Examples

Solved Example

Problem

A = {1,2,3,4}, B = {3,4,5,6} on a die. Find P(A ∪ B).

Solution

P(A ∪ B) = 1 (certain — every outcome is in A or B)

Solved Example

Problem

In a class of 50 students, 30 like Maths and 25 like Science. 15 like both. Find P(student likes at least one).

Solution

P(at least one) = 4/5 = 0.8

Solved Example

Problem

A and B are mutually exclusive with P(A) = 0.35 and P(B) = 0.25. Find P(A ∪ B) and P(neither).

Solution

P(A ∪ B) = 0.60, P(neither) = 0.40

Solved Example

Problem

A and B are mutually exclusive AND exhaustive with P(A) = 0.7. Find P(B).

Solution

P(B) = 0.3

Key Takeaways

Key Takeaways

  • General: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Always works.
  • Mutually exclusive (A ∩ B = ∅): P(A ∪ B) = P(A) + P(B) — no overlap.
  • Mutually exclusive + exhaustive: P(A) + P(B) = 1 — they partition S.
  • Neither: P(neither) = 1 − P(A ∪ B) = P(A' ∩ B').
  • Only one: P(exactly one) = P(A) + P(B) − 2P(A ∩ B).

Practice

  1. P(A) = 0.6, P(B) = 0.4, P(A ∩ B) = 0.2. Find P(A ∪ B) and P(neither).
  2. Two dice are thrown. A = "sum is 6", B = "sum is 8". Find P(A ∪ B). Are they mutually exclusive?
  3. From a deck of 52 cards, P(heart) = 1/4, P(king) = 1/13, P(heart ∩ king) = 1/52. Find P(heart or king).

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