Probability

Addition theorem of probability

The addition theorem of probability quantifies the probability of the union of two or more events, bridging set-theoretic union with measure-theoretic additivity.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

General Addition Theorem of Probability

For any two events $A$ and $B$ defined on a sample space $\Omega$, the probability of their union is given by $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$. For three events $A, B, C$, $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$$.

📐 Geometric Interpretation

Visualized via Venn diagrams, the sum of areas of sets $A$ and $B$ counts their intersection $A \cap B$ twice. Subtracting $P(A \cap B)$ exactly corrects this double-counting, yielding the true area of the union.

⚙️ Algebraic & Structural Meaning

Rooted in the measure-theoretic axioms of Kolmogorov, specifically countable additivity and the inclusion-exclusion principle over Boolean algebras of events.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Event space validity	`Subsets of sample space $\Omega$`	Strictly Mandatory	If $A$ and $B$ are not measurable subsets of $\Omega$, their probabilities are undefined.
Kolmogorov Axioms	$0 \le P(E) \le 1$	Strictly Mandatory	Probabilities outside $[0, 1]$ or violating $P(\Omega) = 1$ invalidate all derived union bounds.
Mutually Exclusive Special Case	$A \cap B = \emptyset$	Optional (reduces formula)	If assumed mutually exclusive when they are not, failure to subtract $P(A \cap B)$ leads to overcounting.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Mutually Exclusive Events

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

Rule: When $A \cap B = \emptyset$, the intersection term vanishes entirely.

Application: undefined

Probability of Neither Event ($A' \cap B'$)

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

Rule: Use De Morgan's Law combined with the addition theorem to find none-occurrences instantly.

Exactly One Event Occurring

$$P(A \Delta B) = P(A) + P(B) - 2P(A \cap B)$$

Rule: Symmetric difference formula accounts for elements in $A$ or $B$ but not both.

Application: Use this result in the stated type of problem.

Bonferroni's First Inequality (Upper/Lower Bounds)

$$P(A \cap B) \ge P(A) + P(B) - 1$$

Rule: Since $P(A \cup B) \le 1$, rearranging the addition theorem gives a strict lower bound for the intersection.

Application: Advanced TGT/PGT questions testing range of intersection given marginal probabilities.

Three-Event Union with Symmetry

$$P(A \cup B \cup C) = 3P(A) - 3P(A \cap B) + P(A \cap B \cap C)$$

Rule: When all individual probabilities and pairwise intersections are equal due to symmetry.

Application: Shortcut for symmetrical die-rolling or card-drawing problems.

Probability of at least two events occurring

$$P(A \cap B) + P(B \cap C) + P(C \cap A) - 2P(A \cap B \cap C)$$

Rule: Rapid calculation trick for multi-event overlap questions.

Application: High-level DSSSB and KVS PGT questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming Independence for General Addition

Common Pitfall: Candidates substitute $P(A \cap B) = P(A)P(B)$ into the addition theorem even when events are dependent.

Counterexample: $A$ and $B$ are dependent events where $P(A \cap B) \neq P(A)P(B)$

Why it fails: Using product formula for intersection on dependent events yields completely wrong union probabilities.

🛡️ Exam Defense: Check explicitly whether independence is stated in the problem before using $P(A)P(B)$.

⚠️ COMMON ERROR #02

Probability Exceeding Unity

Common Pitfall: Calculating $P(A) + P(B)$ and getting a value greater than 1 without realizing $P(A \cap B)$ must absorb the excess.

Counterexample: $P(A) = 0.7, P(B) = 0.6$

Why it fails: Sum is $1.3$. Without subtracting $P(A \cap B)$, the probability axiom $P \le 1$ is violated.

🛡️ Exam Defense: Always verify that $P(A) + P(B) - 1 \le P(A \cap B) \le \min(P(A), P(B))$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
General Addition Theorem	Any two events $A, B$	$P(A \cup B) = P(A) + P(B) - P(A \cap B)$	Union area minus intersection overlap	General overlapping events
Mutually Exclusive Addition	$A \cap B = \emptyset$	$P(A \cup B) = P(A) + P(B)$	Disjoint geometric regions	Keywords: 'either... or...', 'mutually exclusive'
Complementary Probability	Event $A$ and complement $A'$	$P(A) + P(A') = 1$	Total sample space minus $A$	Keywords: 'none', 'neither', 'at least one'
Boole's Inequality	Countable sequence of events $A_n$	$P(\bigcup A_n) \le \sum P(A_n)$	Sum of individual areas exceeds union	Keywords: 'upper bound', 'at least one occurs'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $A$ and $B$ are two events such that $P(A) = 3/8$, $P(B) = 1/2$, and $P(A \cap B) = 1/4$, find $P(A \cup B)$.

(A) 5/8(B) 3/4(C) 1/2(D) 7/8

📚 Method A: Formal Textbook Derivation

Step 1: Write formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Step 2: Substitute values: $3/8 + 1/2 - 1/4 = 3/8 + 4/8 - 2/8 = 5/8$.

⚡ Method B: 25-Second Exam Shortcut

Direct mental arithmetic: $3/8 + 4/8 - 2/8 = 5/8$.

💡 Tip: Convert all fractions to a common denominator instantly.

CANONICAL PROBLEM 02 Dual-Method Breakdown

For two events $A$ and $B$, $P(A) = 0.4$, $P(B) = p$, and $P(A \cup B) = 0.7$. If $A$ and $B$ are mutually exclusive, find the value of $p$.

(A) 0.1(B) 0.2(C) 0.3(D) 0.4

📚 Method A: Formal Textbook Derivation

Step 1: Mutually exclusive means $P(A \cap B) = 0$. Step 2: $P(A \cup B) = P(A) + P(B)$. Step 3: $0.7 = 0.4 + p \implies p = 0.3$.

⚡ Method B: 25-Second Exam Shortcut

Since intersection is zero, simply subtract $P(A)$ from union: $0.7 - 0.4 = 0.3$.

💡 Tip: Spot the keyword 'mutually exclusive' to eliminate the intersection term.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $P(A) = 0.5$, $P(B) = 0.3$, and $P(A \cap B) = 0.1$, then $P(A' \cup B')$ is equal to:

A 0.9

B 0.8

C 0.7

D 0.6

Derivation & Analysis:

$P(A' \cup B') = 1 - P(A \cap B) = 1 - 0.1 = 0.9$ by De Morgan's Law.

MCQ 2: If $A$ and $B$ are mutually exclusive events, then $P(A \cap B)$ is:

A $P(A)P(B)$

B $1$

C $0$

D $P(A) + P(B)$

Derivation & Analysis:

By definition, mutually exclusive events cannot occur simultaneously, so their intersection is the null set, having probability $0$.

MCQ 3: For any two events $A$ and $B$, the probability that exactly one of them occurs is:

A $P(A) + P(B) - P(A \cap B)$

B $P(A) + P(B) - 2P(A \cap B)$

C $P(A) + P(B) + 2P(A \cap B)$

D $P(A \cap B') + P(A' \cap B)$

Derivation & Analysis:

Exactly one occurring is $(A \setminus B) \cup (B \setminus A)$, whose probability is $P(A) + P(B) - 2P(A \cap B)$.

MCQ 4: If $P(A) = 0.6$, $P(B) = 0.4$, and $P(A \cup B) = 0.8$, what is $P(A \cap B)$?

A 0.2

B 0.4

C 0.5

D 0.1

Derivation & Analysis:

$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.4 - 0.8 = 0.2$.

MCQ 5: What is the maximum possible value of $P(A \cap B)$ given $P(A) = 0.3$ and $P(B) = 0.7$?

A 1.0

B 0.7

C 0.3

D 0.21

Derivation & Analysis:

$P(A \cap B) \le \min(P(A), P(B)) = \min(0.3, 0.7) = 0.3$.

MCQ 6: If $A \subset B$, then the value of $P(A \cup B)$ is:

A $P(A)$

B $P(B)$

C $P(A) + P(B)$

D $0$

Derivation & Analysis:

If $A \subset B$, their union is $B$, so $P(A \cup B) = P(B)$.

MCQ 7: If $A$ and $B$ are independent events with $P(A) = 0.4$ and $P(B) = 0.5$, then $P(A \cup B)$ is:

A 0.9

B 0.2

C 0.7

D 0.5

Derivation & Analysis:

For independent events, $P(A \cap B) = 0.4 \times 0.5 = 0.2$. Then $P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7$.

MCQ 8: Assertion (A): For any two events, $P(A \cup B) \ge \max(P(A), P(B))$. Reason (R): Union contains both individual sets as subsets.

A Both (A) and (R) are true and (R) is the correct explanation

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Since $A \subset (A \cup B)$ and $B \subset (A \cup B)$, monotonicity of probability ensures $P(A \cup B) \ge P(A)$ and $P(A \cup B) \ge P(B)$.

MCQ 9: If $P(A) = 1/3$, $P(B) = 1/4$, and $P(A \cup B) = 5/12$, are $A$ and $B$ independent?

A Yes

B No

C Cannot be determined

D Mutually exclusive

Derivation & Analysis:

$P(A \cap B) = 1/3 + 1/4 - 5/12 = 7/12 - 5/12 = 2/12 = 1/6$. Since $P(A)P(B) = (1/3)(1/4) = 1/12 \neq 1/6$, wait... let us recalculate: $1/3 \times 1/4 = 1/12$. Here $P(A \cap B) = 2/12 = 1/6$. Thus they are dependent. Correct option is No (B).

Q10

MCQ 10: In a class of 100 students, 40 failed in Math, 30 failed in Physics, and 10 failed in both. A student is selected at random. What is the probability that the student failed in Math or Physics?

A 0.5

B 0.6

C 0.7

D 0.4

Derivation & Analysis:

Let $M$ be failing Math ($P(M) = 40/100$) and $P$ be failing Physics ($P(P) = 30/100$), $P(M \cap P) = 10/100$. $P(M \cup P) = 0.4 + 0.3 - 0.1 = 0.6$.

Common questions

No. Since $A \subset (A \cup B)$, the probability of the union is always greater than or equal to the probability of any individual constituent event.

Confusing mutually exclusive events with independent events, or incorrectly applying the inclusion-exclusion principle for three or more events.

Mastering Venn diagram allocations and remembering direct formulas for symmetric difference and De Morgan's laws saves crucial seconds.

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