1. Algebra — Probability

Independent events

Two events are said to be independent if the occurrence or non-occurrence of one does not affect the probability of the occurrence or non-occurrence of the other.

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

📖 Topic progress

○ Mark as complete

Definition and statement

Multiplication Rule for Independent Events

Two events $A$ and $B$ associated with a random experiment are said to be independent if $P(A \cap B) = P(A) \cdot P(B)$. More generally, a family of events $(A_i)_{i \in I}$ is mutually independent if for every finite subset $J \subset I$, $P\left(\bigcap_{j \in J} A_j\right) = \prod_{j \in J} P(A_j)$.

📐 Geometric Interpretation

In a sample space represented as a measure space of total area 1, the event $A$ occupies a region of measure $P(A)$ and $B$ occupies $P(B)$. Independence implies that the conditional probability measure of $A$ given $B$ equals the unconditional measure of $A$, meaning $P(A|B) = P(A)$, provided $P(B) > 0$.

⚙️ Algebraic & Structural Meaning

Algebraically, independence simplifies the joint distribution of indicator random variables and provides the foundation for product probability spaces in measure-theoretic probability.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-zero Probabilities for Conditional Form	`Set condition ($P(B) > 0$)`	Strictly Mandatory for conditional probability equivalence	If $P(B) = 0$, the conditional probability $P(A\|B)$ is undefined, though $P(A \cap B) = P(A)P(B)$ still holds trivially as $0 = 0$.
Mutually vs Pairwise Independence	`Finite collection of events ($n \ge 3$)`	Strictly Mandatory for multi-event formulas	Pairwise independence does not imply mutual independence. Assuming mutual independence when only pairwise independence holds leads to false joint probability calculations.
Same Probability Space	`Probability triple $(\Omega, \mathcal{F}, P)$`	Essential for foundational consistency	Events must belong to the same sigma-algebra on the same sample space to evaluate intersection probabilities.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Complementary Independence

$$P(A' \cap B') = P(A')P(B') = (1 - P(A))(1 - P(B))$$

Rule: If $A$ and $B$ are independent, their complements $A'$ and $B'$ are also independent.

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

At Least One Independent Event

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

Rule: To find the probability of occurrence of at least one independent event, subtract the product of probabilities of non-occurrence from 1.

Multi-Event Success Formula

$$P(\text{at least one}) = 1 - \prod_{i=1}^n (1 - P(A_i))$$

Rule: Instantly calculate the probability of at least one success across $n$ independent trials without expanding union formulas.

Application: Directly appears in questions involving multiple shooters or independent components in a system.

Exactly One Event Occurring

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

Rule: For independent events, compute as $P(A)(1-P(B)) + P(B)(1-P(A))$.

Conditional Independence Reduction

$$P(A|B) = P(A) \quad \text{and} \quad P(B|A) = P(B)$$

Rule: If independence is given, eliminate conditional probability bars immediately.

Rapid Elimination via Multiplication

$$P(A \cap B) \neq P(A)P(B) \implies \text{Not Independent}$$

Rule: Check if the given intersection probability matches the product of individual probabilities. If not, instantly eliminate 'independent' options.

Application: Used to quickly screen options in multiple-choice assertion questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusion with Mutually Exclusive Events

Common Pitfall: Candidates assume mutually exclusive events ($A \cap B = \phi$) are independent.

Counterexample: If $P(A) > 0$ and $P(B) > 0$, then $P(A \cap B) = 0 \neq P(A)P(B)$.

Why it fails: Mutually exclusive events with non-zero probability are heavily dependent; if one occurs, the other cannot.

🛡️ Exam Defense: Never confuse 'mutually exclusive' (disjoint sets) with 'independent' (product rule).

⚠️ COMMON ERROR #02

Assuming Pairwise Implies Mutual Independence

Common Pitfall: Assuming that if $P(A_i \cap A_j) = P(A_i)P(A_j)$ for all $i \neq j$, the events are mutually independent.

Counterexample: Bernstein's counterexamples or standard tetrahedral die examples.

Why it fails: Triple intersection condition $P(A \cap B \cap C) = P(A)P(B)P(C)$ may fail even if pairwise intersections hold.

🛡️ Exam Defense: For 3 or more events, verify both pairwise and triple product conditions.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Independent Events	$P(A \cap B) = P(A)P(B)$	$P(A\|B) = P(A)$	Orthogonal/Uncorrelated probability measures	Keywords: 'independently', 'success rate'
Mutually Exclusive Events	$A \cap B = \phi$	$P(A \cup B) = P(A) + P(B)$	Disjoint geometric regions	Keywords: 'either... or', 'cannot occur together'
Conditional Probability	$P(B) > 0$	$P(A\|B) = \frac{P(A \cap B)}{P(B)}$	Restricting sample space to $B$	Keywords: 'given that', 'if it is known that'
Exhaustive Events	$\bigcup_{i=1}^n A_i = \Omega$	$\sum_{i=1}^n P(A_i) = 1$	Covering the entire sample space	Keywords: 'form a partition', 'comprehensive set'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $A$ and $B$ are two independent events such that $P(A) = 0.3$ and $P(A \cup B) = 0.8$, find $P(B)$.

(A) $\frac{5}{7}$(B) $\frac{2}{7}$(C) $\frac{3}{7}$(D) $\frac{4}{7}$

📚 Method A: Formal Textbook Derivation

Step 1: Use the union formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Step 2: Since $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$. Step 3: Substitute values: $0.8 = 0.3 + P(B) - 0.3P(B)$. Step 4: $0.5 = 0.7P(B) \implies P(B) = \frac{0.5}{0.7} = \frac{5}{7}$.

⚡ Method B: 25-Second Exam Shortcut

Use $1 - P(A \cup B) = (1 - P(A))(1 - P(B)) \implies 0.2 = 0.7(1 - P(B)) \implies 1 - P(B) = \frac{2}{7} \implies P(B) = \frac{5}{7}$.

💡 Tip: Always convert union to complement product for independent events to save calculation time.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The probability that a student solves a mathematics problem is $\frac{1}{2}$ and another student solves it is $\frac{1}{3}$. If they try independently, what is the probability that the problem is solved?

(A) $\frac{1}{6}$(B) $\frac{2}{3}$(C) $\frac{1}{3}$(D) $\frac{5}{6}$

📚 Method A: Formal Textbook Derivation

Step 1: Let $A$ and $B$ be the events. $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$. Step 2: Probability neither solves is $P(A')P(B') = (1 - \frac{1}{2})(1 - \frac{1}{3}) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$. Step 3: Probability problem is solved = $1 - \frac{1}{3} = \frac{2}{3}$.

⚡ Method B: 25-Second Exam Shortcut

P(At least one) = $1 - (1 - \frac{1}{2})(1 - \frac{1}{3}) = 1 - \frac{1}{3} = \frac{2}{3}$.

💡 Tip: Recognize 'problem is solved' as 'at least one person solves it'.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $A$ and $B$ are independent events with $P(A) = 0.2$ and $P(B) = 0.5$, then $P(A \cup B)$ is equal to:

A 0.7

B 0.6

C 0.1

D 0.3

Derivation & Analysis:

$P(A \cup B) = P(A) + P(B) - P(A)P(B) = 0.2 + 0.5 - (0.2)(0.5) = 0.7 - 0.1 = 0.6$.

MCQ 2: If $A$ and $B$ are independent events such that $P(A) = \frac{3}{10}$ and $P(A \cup B) = \frac{8}{10}$, then $P(B)$ is:

A $\frac{5}{7}$

B $\frac{2}{7}$

C $\frac{3}{7}$

D $\frac{4}{7}$

Derivation & Analysis:

$P(A \cup B) = P(A) + P(B) - P(A)P(B) \implies \frac{8}{10} = \frac{3}{10} + P(B)(1 - \frac{3}{10}) \implies \frac{5}{10} = \frac{7}{10}P(B) \implies P(B) = \frac{5}{7}$.

MCQ 3: If $A$ and $B$ are two independent events, then $P(A \text{ alone})$ is given by:

A $P(A) - P(B)$

B $P(A)(1 - P(B))$

C $P(A)P(B)$

D $1 - P(A)$

Derivation & Analysis:

$P(A \cap B') = P(A)P(B') = P(A)(1 - P(B))$ represents the probability that event $A$ occurs and event $B$ does not occur.

MCQ 4: If $P(A) = 0.4$, $P(B) = x$, and $A, B$ are independent with $P(A \cup B) = 0.7$, the value of $x$ is:

A 0.3

B 0.4

C 0.5

D 0.6

Derivation & Analysis:

$0.7 = 0.4 + x - 0.4x \implies 0.3 = 0.6x \implies x = 0.5$.

MCQ 5: Which of the following statements is always true for independent events $A$ and $B$ with non-zero probabilities?

A $A$ and $B$ are mutually exclusive

B $A$ and $B'$ are independent

C $P(A|B) = 0$

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

Derivation & Analysis:

If $A$ and $B$ are independent, any combination of $A$ or $A'$ and $B$ or $B'$ are also independent. Thus $A$ and $B'$ are independent.

MCQ 6: Three independent shooters hit a target with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$. The probability that the target is hit is:

A $\frac{1}{24}$

B $\frac{3}{4}$

C $\frac{1}{4}$

D $\frac{23}{24}$

Derivation & Analysis:

$P(\text{hit}) = 1 - P(\text{none hit}) = 1 - (1 - \frac{1}{2})(1 - \frac{1}{3})(1 - \frac{1}{4}) = 1 - (\frac{1}{2})(\frac{2}{3})(\frac{3}{4}) = 1 - \frac{1}{4} = \frac{3}{4}$.

MCQ 7: If $A$ and $B$ are independent events, then $P(A \cup B | C)$ equals (assuming $P(C) > 0$):

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

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

C $P(A)P(B)$

D 0

Derivation & Analysis:

Conditional probability preserves independence given a conditioning event $C$ under standard compatibility, yielding the standard union expansion in conditional form.

MCQ 8: Assertion (A): If $A$ and $B$ are independent, then $P(A \cap B) = P(A)P(B)$. Reason (R): Independence means the occurrence of one does not depend on the other.

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

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:

Both statements are mathematically rigorous definitions and correctly linked.

MCQ 9: If $P(A) = 0.3, P(B) = 0.4$, and $P(A \cap B) = 0.12$, then $A$ and $B$ are:

A Mutually exclusive

B Independent

C Dependent

D Exhaustive

Derivation & Analysis:

Since $P(A)P(B) = 0.3 \times 0.4 = 0.12 = P(A \cap B)$, the events are independent.

Q10

MCQ 10: If $A$ and $B$ are independent events such that $P(A) = p, P(B) = q$, and $P(A' \cap B') = \frac{1}{4}$, then:

A $pq = \frac{3}{4}$

B $(1-p)(1-q) = \frac{1}{4}$

C $p+q = \frac{3}{4}$

D $p+q - pq = \frac{3}{4}$

Derivation & Analysis:

By independence of complements, $P(A' \cap B') = P(A')P(B') = (1-p)(1-q) = \frac{1}{4}$.

Common questions

Yes, but only if at least one of the events has a probability of zero. If $P(A) > 0$ and $P(B) > 0$, mutual exclusivity ($P(A \cap B) = 0$) and independence ($P(A)P(B) > 0$) are mutually exclusive.

Confusing mutual exclusivity with independence, or failing to check triple intersection conditions when dealing with three or more events.

Always utilize complement formulas ($1 - P(A')$) and multiplicative rules for joint occurrences to bypass lengthy Venn diagram constructions.

← Previous Topic

Conditional probability

Next Topic →

Multiplication theorem of probability