Law of Addition of Probability — Class 12 Statistics Notes (GSEB)

The Law — Overlap Correction

The rule of obtaining the probability of the occurrence of at least one of the events A and B in the sample space of a random experiment is called the law of addition of probability. Since “at least one of A and B” is A ∪ B, this law is the rule for P(A ∪ B). It is stated as follows and accepted without proof: ⭐

See the law emerge — multiples of 2 or 3 from the first 50 natural numbers

A = multiples of 2 (25) · B = multiples of 3 (16) · A ∩ B = multiples of 6 (8). Count each region, then compare with the formula.

First 50 natural numbers

A = multiples of 2 (25) · B = multiples of 3 (16) · A ∩ B = multiples of 6 (8)

n(U) = 50

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A only B only A ∩ B outside

Probability

Click a region button to shade it and read its count using the controls above.

Law of Addition of Probability

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

Addition Calculator — type fractions or decimals, watch the union come out

P(A)P(B)P(A ∩ B)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

= 2/3≈ 0.6667

Three-Event Law and Derived Results

Results (3) and (4) are the engine behind “P(A) = 2P(B), find P(A)” style questions (Illustrations 25, 26; Exercises 1.3 Q9–10).

Law of Addition — three events ⭐

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)

Result (1) — A and B mutually exclusive ⭐

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

Result (2) — A, B, C mutually exclusive

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

Result (3) — A and B mutually exclusive and exhaustive ⭐

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

Result (4) — A, B, C mutually exclusive and exhaustive ⭐

P(A \cup B \cup C) = P(A) + P(B) + P(C) = 1

Solved Illustrations 19–26

Step 1 — the sample space

1/5

One number is selected from the first 50 natural numbers: n = ⁵⁰C₁ = 50.

Solved Example

Problem

Illustration 20: one card is drawn from a pack of 52. A = club card, B = queen. Find the probability that the card is (1) a club or a queen (2) neither a club nor a queen.

Solution

(1) 4/13 (2) 9/13.

The 16 cards of A ∪ B — see the overlap

Suit ↓ / Rank →	A	2	3	4	5	6	7	8	9	10	J	Q	K
♠													♠K
♦													♦K
♣													♣K
♥	♥A	♥2	♥3	♥4	♥5	♥6	♥7	♥8	♥9	♥10	♥J	♥Q	♥K

13 hearts (A) + 4 kings (B) − 1 heart king (A ∩ B) = 16 highlighted cells = A ∪ B.

Solved Example

Problem

Illustration 21: 3 medical and 5 engineering persons are available; 2 are selected for a committee. Find the probability that both selected are of the same profession.

Solution

Required probability = 13/28.

Step 1 — reads at least one

1/3

P(A) = 0.55 (reads newspaper X), P(B) = 0.69 (reads newspaper Y), P(A ∩ B) = 0.27.

P(A ∪ B) = 0.55 + 0.69 − 0.27 = 0.97

Solved Example

Problem

Illustration 23 (⭐ board): P(A) = 2P(B) = 4P(A ∩ B) = 0.6. Find (1) P(A′ ∩ B′) (2) P(A′ ∪ B′) (3) P(A − B) (4) P(B − A).

Solution

(1) 0.25 (2) 0.85 (3) 0.45 (4) 0.15.

Solved Example

Problem

Illustration 24: P(A′) = 0.3, P(B) = 0.6 and P(A ∪ B) = 0.83. Find P(A ∩ B), P(A ∩ B′) and P(A′ ∩ B).

Solution

P(A ∩ B) = 0.47; P(A ∩ B′) = 0.23; P(A′ ∩ B) = 0.13.

Solved Example

Problem

Illustration 25: 3P(A) = 4P(B) = 1 and A, B are mutually exclusive. Find P(A ∪ B).

Solution

Required probability = 7/12.

Solved Example

Problem

Illustration 26 (⭐ board): 2P(A) = 3P(B) = 4P(C) and A, B, C are mutually exclusive and exhaustive. Find P(A ∪ B) and P(B ∪ C).

Solution

P(A ∪ B) = 10/13; P(B ∪ C) = 7/13.

Practice Checkpoint

Your turn. Work each question in your notebook — type only the final answer here.

Addition Q6

3 marks

A card is randomly selected from a pile of 52 cards. Find the probability that it is (1) a club or a queen (2) neither a club nor a queen (3) a spade or an ace (4) neither a spade nor an ace.

Try the experiment:

♠

♣

♥

♦

The drawn card light up in the 52-card deck. Draws are with replacement (the deck is unchanged).

Addition Q1 (Board)

⭐ Board

2 marks

Two balanced dice are tossed together. Find the probability that the sum of the digits on both sides is a multiple of 2 or 3.

Try the experiment:

1,1

1,2

1,3

1,4

1,5

1,6

2,1

2,2

2,3

2,4

2,5

2,6

3,1

3,2

3,3

3,4

3,5

3,6

4,1

4,2

4,3

4,4

4,5

4,6

5,1

5,2

5,3

5,4

5,5

5,6

6,1

6,2

6,3

6,4

6,5

6,6

Throw the dice — the outcome lights up in the 6 × 6 grid.

Addition Q25

2 marks

A and B are mutually exclusive and exhaustive events in a sample space and P(A) = 2P(B). Find P(A).

Finish the rest of this question type → Practice Page· then return here for the next topic

Key Takeaways

The addition law: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) — the intersection is subtracted because it was counted twice.
Three events: add all three, subtract the three pairwise intersections, add back the triple intersection.
Mutually exclusive → P(A ∪ B) = P(A) + P(B) (the overlap term collapses to zero).
Mutually exclusive and exhaustive → P(A ∪ B) = P(A) + P(B) = 1.
Only one of A or B = (A ∩ B′) ∪ (A′ ∩ B) with P = [P(A) − P(A ∩ B)] + [P(B) − P(A ∩ B)].
Neither A nor B = A′ ∩ B′ = (A ∪ B)′ → 1 − P(A ∪ B).