2026?27 Syllabus Focus
The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.
- Conditional probability and the multiplication theorem.
- Independent events.
- Total probability and Bayes' theorem.
⏱️ Last 24-Hour Checklist
Key Formulas (Must Know!)
Key Concepts
Conditional Probability ★★★★★
Definition: The conditional probability of event E given that F has occurred
is:
$$P(E|F) = \frac{P(E \cap F)}{P(F)}, \quad P(F) \neq 0$$
Read as: "Probability of E given F"
Understanding Conditional Probability
When event F has occurred, the sample space reduces from S to F. We now find the probability of E within this reduced sample space.
$$P(E|F) = \frac{Number of outcomes favorable to both E and F}{Number of outcomes
favorable to F}$$
$$= \frac{n(E \cap F)}{n(F)} = \frac{P(E \cap F)}{P(F)}$$
Properties of Conditional Probability
| Property | Formula |
|---|---|
| Property 1 | $P(S|F) = P(F|F) = 1$ |
| Property 2 | $P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$ |
| Property 2 (Disjoint) | If A and B are disjoint: $P((A \cup B)|F) = P(A|F) + P(B|F)$ |
| Property 3 | $P(E'|F) = 1 - P(E|F)$ |
📌 Key Point: Conditional probability changes the sample space. Always think:
"Given F has occurred, what's the probability of E within F?"
🔍 Simple Example: In a family with 2 children, if we know at least one is a boy, the sample space reduces from {BB, BG, GB, GG} to just {BB, BG, GB}. Now finding P(both boys) = 1/3, not 1/4!
🔍 Simple Example: In a family with 2 children, if we know at least one is a boy, the sample space reduces from {BB, BG, GB, GG} to just {BB, BG, GB}. Now finding P(both boys) = 1/3, not 1/4!
Multiplication Theorem ★★★★☆
Multiplication Rule:
$$P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F)$$
Probability of both E and F occurring = Probability of one × Conditional probability of other
For More Than Two Events
$$P(E \cap F \cap G) = P(E) \cdot P(F|E) \cdot P(G|EF)$$
This extends to any number of events
When to Use Multiplication Theorem
- Drawing cards/balls without replacement
- Sequential events where outcome of first affects second
- Finding probability of intersection of events
- Problems involving "and" (both events occurring)
Classic Example: Two cards are drawn from deck without replacement. Find P(both
are kings).
Solution:
Let E = first card is king, F = second card is king
$P(E) = \frac{4}{52}$
$P(F|E) = \frac{3}{51}$ (3 kings left in 51 cards)
$P(E \cap F) = P(E) \cdot P(F|E) = \frac{4}{52} × \frac{3}{51} = \frac{1}{221}$
Solution:
Let E = first card is king, F = second card is king
$P(E) = \frac{4}{52}$
$P(F|E) = \frac{3}{51}$ (3 kings left in 51 cards)
$P(E \cap F) = P(E) \cdot P(F|E) = \frac{4}{52} × \frac{3}{51} = \frac{1}{221}$
Independent Events ★★★★★
Definition: Two events E and F are independent if:
$$P(E \cap F) = P(E) \cdot P(F)$$
Equivalently: $P(E|F) = P(E)$ and $P(F|E) = P(F)$
Meaning: Occurrence of one event does NOT affect probability of the other
Independent vs Mutually Exclusive
| Aspect | Independent Events | Mutually Exclusive Events |
|---|---|---|
| Definition | $P(E \cap F) = P(E) \cdot P(F)$ | $E \cap F = φ$ (cannot occur together) |
| Occurrence | Can occur simultaneously | Cannot occur simultaneously |
| Effect | One doesn't affect other | One rules out the other |
| $P(E \cap F)$ | $P(E) × P(F)$ (may be > 0) | 0 (always zero) |
| Example | Tossing two coins independently | Getting H and T on same coin toss |
⚠️ Most Common Board Exam Mistake: Confusing "independent" with "mutually
exclusive"!
Remember:
- Independent = Events can happen together, one doesn't affect the other (multiply probabilities)
- Mutually Exclusive = Events CANNOT happen together (P(EnF) = 0)
Easy Memory Trick:
- INdependent → INtersection exists, probabilities multiply
- EXclusive → EXcludes each other, intersection is empty
Remember:
- Independent = Events can happen together, one doesn't affect the other (multiply probabilities)
- Mutually Exclusive = Events CANNOT happen together (P(EnF) = 0)
Easy Memory Trick:
- INdependent → INtersection exists, probabilities multiply
- EXclusive → EXcludes each other, intersection is empty
Properties of Independent Events
If E and F are independent, then:
- E and F' are independent
- E' and F are independent
- E' and F' are independent
- E and F' are independent
- E' and F are independent
- E' and F' are independent
Three Events Independence
Events A, B, C are mutually independent if ALL of the following hold:
- $P(A \cap B) = P(A) \cdot P(B)$
- $P(A \cap C) = P(A) \cdot P(C)$
- $P(B \cap C) = P(B) \cdot P(C)$
- $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$
- $P(A \cap B) = P(A) \cdot P(B)$
- $P(A \cap C) = P(A) \cdot P(C)$
- $P(B \cap C) = P(B) \cdot P(C)$
- $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$
Partition of Sample Space & Total Probability ★★★★☆
Partition of Sample Space:
Events $E_1, E_2, ..., E_n$ form a partition of S if:
1. $E_i \cap E_j = φ$ for $i \neq j$ (pairwise disjoint)
2. $E_1 \cup E_2 \cup ... \cup E_n = S$ (exhaustive)
3. $P(E_i) > 0$ for all i (non-zero probability)
Events $E_1, E_2, ..., E_n$ form a partition of S if:
1. $E_i \cap E_j = φ$ for $i \neq j$ (pairwise disjoint)
2. $E_1 \cup E_2 \cup ... \cup E_n = S$ (exhaustive)
3. $P(E_i) > 0$ for all i (non-zero probability)
Theorem of Total Probability
If $\{E_1, E_2, ..., E_n\}$ is a partition of S and A is any event, then:
$$P(A) = ∑_{i=1}^{n} P(E_i) \cdot P(A|E_i)$$
$$= P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + ... + P(E_n)P(A|E_n)$$
Use Case: When you can't find P(A) directly, but you know:
- Different ways A can occur (through $E_1, E_2, ..., E_n$)
- Probability of each way: $P(E_i)$
- Conditional probability: $P(A|E_i)$ for each way
- Different ways A can occur (through $E_1, E_2, ..., E_n$)
- Probability of each way: $P(E_i)$
- Conditional probability: $P(A|E_i)$ for each way
Example: Two bags: Bag I has 3 red, 4 black balls. Bag II has 5 red, 6 black
balls. A bag is chosen randomly and a ball drawn. Find P(red ball).
Solution:
Let $E_1$ = Bag I chosen, $E_2$ = Bag II chosen, A = red ball drawn
$P(E_1) = P(E_2) = \frac{1}{2}$
$P(A|E_1) = \frac{3}{7}$, $P(A|E_2) = \frac{5}{11}$
By Total Probability:
$P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)$
$= \frac{1}{2} × \frac{3}{7} + \frac{1}{2} × \frac{5}{11} = \frac{3}{14} + \frac{5}{22} = \frac{34}{77}$
Solution:
Let $E_1$ = Bag I chosen, $E_2$ = Bag II chosen, A = red ball drawn
$P(E_1) = P(E_2) = \frac{1}{2}$
$P(A|E_1) = \frac{3}{7}$, $P(A|E_2) = \frac{5}{11}$
By Total Probability:
$P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)$
$= \frac{1}{2} × \frac{3}{7} + \frac{1}{2} × \frac{5}{11} = \frac{3}{14} + \frac{5}{22} = \frac{34}{77}$
Bayes' Theorem
Bayes' Theorem:
If $\{E_1, E_2, ..., E_n\}$ is a partition of S and A is any event with $P(A) \neq 0$, then: $P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{∑_{j=1}^{n} P(E_j) \cdot P(A|E_j)}$
For two events: $P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
If $\{E_1, E_2, ..., E_n\}$ is a partition of S and A is any event with $P(A) \neq 0$, then: $P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{∑_{j=1}^{n} P(E_j) \cdot P(A|E_j)}$
For two events: $P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
Understanding Bayes' Theorem
Purpose: Find "reverse" probability - if we know event A has already occurred, what's the probability it was caused by hypothesis $E_i$?
Tree Diagram for Total Probability & Bayes' Theorem
Total Probability: $P(A) = Sum of top and bottom purple branches$.
Bayes' Theorem: $P(E_1|A) = \frac{Specific Purple Branch}{Sum of all Purple Branches}$.
| Term | Meaning | Symbol |
|---|---|---|
| Hypotheses/Causes | Different possible causes or sources | $E_1, E_2, ..., E_n$ |
| Prior Probability | Probability of hypothesis before evidence | $P(E_i)$ |
| Likelihood | Probability of evidence given hypothesis | $P(A|E_i)$ |
| Posterior Probability | Probability of hypothesis after evidence | $P(E_i|A)$ |
Step-by-Step Approach
Step 1: Identify the hypotheses (causes) $E_1, E_2, ..., E_n$
Step 2: Find prior probabilities $P(E_i)$
Step 3: Find likelihoods $P(A|E_i)$ for each hypothesis
Step 4: Apply Bayes' formula
Step 5: Simplify and calculate
Step 2: Find prior probabilities $P(E_i)$
Step 3: Find likelihoods $P(A|E_i)$ for each hypothesis
Step 4: Apply Bayes' formula
Step 5: Simplify and calculate
🎯 When to Use Bayes' Theorem:
Keywords to Look For:
- "What's the probability it came from..."
- "Given that [result], find probability of [cause]"
- "If we know [evidence], what's P([source])"
- "After observing [result], find P([origin])"
Common Question Types:
- Medical: Given positive test, P(actually has disease)?
- Manufacturing: Given defective item, P(from machine A)?
- Bags/Urns: Given red ball drawn, P(from Bag I)?
- Quality: Given item is good, P(from supplier X)?
💡 The Key Difference:
- Regular conditional: P(Result | Known Cause)
- Bayes': P(Unknown Cause | Known Result) → Reverse!
Keywords to Look For:
- "What's the probability it came from..."
- "Given that [result], find probability of [cause]"
- "If we know [evidence], what's P([source])"
- "After observing [result], find P([origin])"
Common Question Types:
- Medical: Given positive test, P(actually has disease)?
- Manufacturing: Given defective item, P(from machine A)?
- Bags/Urns: Given red ball drawn, P(from Bag I)?
- Quality: Given item is good, P(from supplier X)?
💡 The Key Difference:
- Regular conditional: P(Result | Known Cause)
- Bayes': P(Unknown Cause | Known Result) → Reverse!
Random Variables & Probability Distributions ★★★★☆
Random Variable (X): A real valued function whose domain is the sample space of a
random experiment.
Probability Distribution: The system of numbers mapping values $x_i$ to their probabilities $P(X=x_i) = p_i$.
Condition: $∑ p_i = 1$ and $p_i ≥ 0$.
Probability Distribution: The system of numbers mapping values $x_i$ to their probabilities $P(X=x_i) = p_i$.
Condition: $∑ p_i = 1$ and $p_i ≥ 0$.
Mean (Expectation) $μ$ or $E(X)$:
$$E(X) = ∑_{i=1}^{n} x_i p_i$$
⚡ High-Yield Probability Theorems & Distribution Rules (CBSE 4/5 Marks):
- Independent Events Fundamental Identities (MCQ & A-R Classic):
- $A$ and $B$ are independent $\iff P(A \cap B) = P(A) \cdot P(B)$.
- If $A$ and $B$ are independent, then:
- $A$ and $B'$ are independent.
- $A'$ and $B$ are independent.
- $A'$ and $B'$ are independent $\implies \mathbf{P(A' \cap B') = P(A') \cdot P(B') = (1 - P(A))(1 - P(B))}$.
- At least one occurs: $P(A \cup B) = 1 - P(A' \cap B') = 1 - (1 - P(A))(1 - P(B))$.
- Mutually Exclusive vs. Independent: Two events with positive probabilities CANNOT be both mutually exclusive ($P(A \cap B) = 0$) and independent ($P(A \cap B) = P(A)P(B) > 0$).
- Conditional Probability Properties:
- $P(A|B) = \frac{P(A \cap B)}{P(B)}$ ($P(B) \neq 0$)
- $P(S|B) = 1$ and $P(B|B) = 1$
- Complementary Event: $P(A'|B) = 1 - P(A|B)$
- $P((A \cup B)|E) = P(A|E) + P(B|E) - P((A \cap B)|E)$
- Theorem of Total Probability & Bayes' Theorem:
- Let $\{E_1, E_2, \dots, E_n\}$ be a partition of sample space $S$ (pairwise disjoint with $\sum P(E_i) = 1$).
- Total Probability: $P(A) = \sum_{i=1}^n P(E_i) \cdot P(A|E_i)$.
- Bayes' Theorem:
$$P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{\sum_{j=1}^n P(E_j) \cdot P(A|E_j)}$$
- Random Variable & Probability Distribution Properties:
- For any valid probability distribution: $p_i \ge 0$ and $\sum p_i = 1$.
- Finding Unknown Parameter $k$: Equate $\sum P(X=x_i) = 1$ to solve for $k$. Reject negative roots or values where $p_i < 0$.
- Mean (Expected Value) of $X$:
$$\mu = E(X) = \sum x_i p_i$$
- 📌 Syllabus Clarification: Variance of random variable $\text{Var}(X) = \sum x_i^2 p_i - \mu^2$, standard deviation, and Bernoulli Trials & Binomial Distribution (NCERT Ex 13.5) have been rationalized and deleted from the CBSE Class 12 Board syllabus.
Solved Examples (Board Marking Scheme)
Example 1: Conditional Probability (2 Marks)
Question: If P(A) = 0.6, P(B) = 0.3 and P(A∩B) = 0.2, find P(A|B) and P(B|A).
Solution: 2 Marks
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} = \frac{2}{3}$
$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.6} = \frac{1}{3}$
Example 2: Multiplication Rule (3 Marks)
Question: An urn contains 10 black and 5 white balls. Two balls are drawn without replacement. Find probability both are black.
Solution: 3 Marks
Let E = first ball is black, F = second ball is black
$P(E) = \frac{10}{15} = \frac{2}{3}$
After drawing one black ball: $P(F|E) = \frac{9}{14}$
$P(E \cap F) = P(E) \cdot P(F|E) = \frac{2}{3} × \frac{9}{14} = \frac{3}{7}$
Example 3: Bayes' Theorem (5 Marks)
Question: Bag I: 3 red, 4 black. Bag II: 5 red, 6 black. A ball is drawn from a random bag and is red. Find P(from Bag II).
Solution: 5 Marks
$E_1$ = Bag I, $E_2$ = Bag II, A = red ball
$P(E_1) = P(E_2) = \frac{1}{2}$
$P(A|E_1) = \frac{3}{7}$, $P(A|E_2) = \frac{5}{11}$
$P(E_2|A) = \frac{\frac{1}{2} × \frac{5}{11}}{\frac{1}{2} × \frac{3}{7} + \frac{1}{2} × \frac{5}{11}} = \frac{\frac{5}{22}}{\frac{68}{154}} = \frac{35}{68}$
Previous Year Questions (PYQs)
2023 (1M MCQ): If P(A) = 0.5, P(B) = 0, then P(A|B) is
(A) 0 (B) 0.5 (C) not defined (D) 1
Answer: (C) - P(B) = 0, so division by zero → not defined
(A) 0 (B) 0.5 (C) not defined (D) 1
Answer: (C) - P(B) = 0, so division by zero → not defined
2022 (3M): Family has two children. Find P(both girls | (i) youngest is girl
(ii) at least one is girl).
Solution: S = {BB, BG, GB, GG}
(i) F = {BG, GG} → P = 1/2
(ii) F = {BG, GB, GG} → P = 1/3
Solution: S = {BB, BG, GB, GG}
(i) F = {BG, GG} → P = 1/2
(ii) F = {BG, GB, GG} → P = 1/3
2021 (5M): Factory: A (25%), B (35%), C (40%). Defects: 5%, 4%, 2%. A defective
bolt is found. P(from B)?
Answer: Using Bayes': $\frac{0.35 × 0.04}{0.25 × 0.05 + 0.35 × 0.04 + 0.40 × 0.02} = \frac{28}{69}$
Answer: Using Bayes': $\frac{0.35 × 0.04}{0.25 × 0.05 + 0.35 × 0.04 + 0.40 × 0.02} = \frac{28}{69}$
Common Mistakes & Scoring Tips
Common Mistakes ⚠️
Mistake 1: Confusing P(E|F) with P(F|E)
They're NOT the same! P(rain|clouds) ≠ P(clouds|rain)
They're NOT the same! P(rain|clouds) ≠ P(clouds|rain)
Mistake 2: Thinking Independent = Mutually Exclusive
Opposite concepts! Independent CAN occur together, Mutually Exclusive CANNOT
Opposite concepts! Independent CAN occur together, Mutually Exclusive CANNOT
Mistake 3: Forgetting to update counts in "without replacement"
If you draw 1 red ball from 5, next time you have only 4 balls left!
If you draw 1 red ball from 5, next time you have only 4 balls left!
Mistake 4: Missing terms in Bayes' denominator
Must include ALL hypotheses: P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + ...
Must include ALL hypotheses: P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + ...
Mistake 5: Not checking if P(F) = 0
Can't divide by zero! P(E|F) undefined if P(F) = 0
Can't divide by zero! P(E|F) undefined if P(F) = 0
Mistake 6: Writing P(E∩F) for independent events without verifying
First check: $P(E) \cdot P(F) = P(E \cap F)$? Only then are they independent!
First check: $P(E) \cdot P(F) = P(E \cap F)$? Only then are they independent!
Scoring Tips 🎯
Tip 1: Define events first
"Let E = ..., F = ..." gets you 0.5 marks even if rest is wrong!
"Let E = ..., F = ..." gets you 0.5 marks even if rest is wrong!
Tip 2: Draw tree diagrams
Visual = fewer mistakes + shows understanding = more marks
Visual = fewer mistakes + shows understanding = more marks
Tip 3: Write the formula before substituting
Shows you know the concept, gets partial marks even if calculation wrong
Shows you know the concept, gets partial marks even if calculation wrong
Tip 4: Show intermediate steps
Don't jump to answer! Each step = marks
Don't jump to answer! Each step = marks
Tip 5: Simplify fractions completely
3/6 should be written as 1/2 for full marks
3/6 should be written as 1/2 for full marks
Tip 6: Box or underline final answer
Makes examiner's job easy = happy examiner = generous marking!
Makes examiner's job easy = happy examiner = generous marking!
⚡ Quick Success Formula
For 10/10 in Probability:1. Memorize 6 formulas (conditional, multiplication, independence, total prob, Bayes, P(E'|F))
2. Solve 15 PYQs (covers all question types)
3. Practice drawing tree diagrams
4. Understand difference: Independent vs Mutually Exclusive
5. Master Bayes' theorem pattern recognition
Time Allocation in Exam:
- 1M MCQ: 1 minute
- 2M questions: 3-4 minutes
- 3M questions: 5-6 minutes
- 5M questions: 8-10 minutes
Practice Problems
Level 1: Basic (2M)
Q1. P(E) = 0.6, P(F) = 0.3, P(E∩F) = 0.2. Find P(E|F) and P(F|E).
Answer: P(E|F) = 2/3, P(F|E) = 1/3
Q2. If A and B independent with P(A) = 3/5, P(B) = 1/5, find P(A∩B).
Answer: 3/25
Level 2: Intermediate (3M)
Q3. Box 1: 4 red, 4 black. Box 2: 2 red, 6 black. Random box chosen, red ball drawn. P(from Box 1)?
Answer: 2/3
Level 3: Advanced (5M)
Q4. 60% hostellers, 40% day scholars. 30% hostellers get A grade, 20% day scholars get A. Student has A grade. P(hosteller)?
Answer: 9/13
Quick Reference Formulas
1. Conditional Probability:
$P(E|F) = \frac{P(E \cap F)}{P(F)}$
2. Multiplication:
$P(E \cap F) = P(E) \cdot P(F|E)$
3. Independence:
$P(E \cap F) = P(E) \cdot P(F)$
4. Total Probability:
$P(A) = ∑ P(E_i) \cdot P(A|E_i)$
5. Bayes' Theorem:
$P(E_i|A) = \frac{P(E_i)P(A|E_i)}{∑ P(E_j)P(A|E_j)}$
When to Use:
"Given that" → Conditional | "Both" → Multiplication
"Independent" → Check/Use Independence | "Which source" → Bayes'
"Given that" → Conditional | "Both" → Multiplication
"Independent" → Check/Use Independence | "Which source" → Bayes'
Final Checklist
📋 Before Exam
- Conditional probability formula
- Multiplication theorem
- Independence condition
- Total probability theorem
- Bayes' theorem
- P(E'|F) = 1 - P(E|F)
✍️ During Exam
- Define events clearly
- Check P(F) ≠ 0 before using P(E|F)
- Show all steps in calculations
- Draw tree diagrams for sequential events
- Verify independence before assuming
- Include ALL terms in Bayes' denominator
Golden Rules
1. Define events: "Let E = ..., F = ..."2. Write formula before substituting
3. Show all steps clearly
4. Simplify fractions
5. Check: $0 ≤ P(E) ≤ 1$
6. Box final answer
Chapter 13 Board Mastery Test
Practice test • 10 marks • 25 min
Q1 (1 Mark): If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find $P(A|B)$.
Q2 (2 Marks): Two cards are drawn without replacement from a well-shuffled pack of 52 cards. Find the probability that both cards are black.
Q3 (3 Marks): A random variable $X$ has the following probability distribution:
Find the value of $k$ and evaluate $P(1 ≤ X < 3)$.
| $X$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(X)$ | $k$ | $2k$ | $3k$ | $4k$ |
Q4 (4 Marks): Bag I contains 3 red and 4 black balls and Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from Bag II.