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CBSE Class 12 Mathematics★ 8-10 marks · CBSE Class 12 Mathematics Chapter 13

Probability

The final chapter of Class 12 Mathematics deepens probability: conditional probability, the Multiplication Theorem, Bayes' Theorem, random variables, and probability distributions (binomial). CBSE Class 12 Mathematics Chapter 13.

⏱ 22 min readHard difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Compute conditional probability P(A|B) = P(A∩B)/P(B) and use the multiplication theorem
2Determine whether events are independent using P(A∩B) = P(A)·P(B)
3Apply the Total Probability Theorem to find P(A) from a partition of the sample space
4Apply Bayes' Theorem to find P(Eᵢ|A) given prior probabilities and likelihoods
5Define a random variable X, construct its probability distribution, and compute mean E(X) = Σxᵢpᵢ and variance Var(X) = E(X²) − [E(X)]²; apply the Binomial distribution

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Why this chapter matters

Probability is the second-highest-weightage chapter in Class 12 Mathematics after Integrals. Bayes' Theorem is guaranteed in every board exam. Binomial distribution problems (mean = np, variance = npq) are the reliable 3-mark question. Conditional probability and independence are tested as MCQs and short-answer questions every year.

Probability — Class 12

"Probability is the measure of our ignorance. The better we understand randomness, the better we can make decisions under uncertainty."

1. Chapter Overview

Building on Class 11, this chapter takes probability to an ADVANCED level: conditional probability (P(A|B) — the probability of A GIVEN that B has occurred), Multiplication Theorem, the Law of Total Probability, Bayes' Theorem (the most important theorem in the chapter), random variables (discrete), probability distributions, and the Binomial distribution.

2. Conditional Probability

P(A|B) = P(A ∩ B) / P(B). Provided P(B) > 0.
'The probability of A GIVEN that B has occurred.'
Multiplication Theorem: P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
Independent Events: A and B are independent if P(A ∩ B) = P(A) × P(B). OR: P(A|B) = P(A) (B gives no information about A).

3. Law of Total Probability

If B₁, B₂, ..., Bₙ are MUTUALLY EXCLUSIVE and EXHAUSTIVE: P(A) = P(B₁)P(A|B₁) + P(B₂)P(A|B₂) + ... + P(Bₙ)P(A|Bₙ)

4. Bayes' Theorem — THE Key Result

$P (B_{i} ∣ A) = \frac{P ( B _{i} ) \cdot P ( A ∣ B _{i} )}{\sum _{j = 1}^{n} P ( B _{j} ) \cdot P ( A ∣ B _{j} )}$

'Bayes' Theorem REVERSES the conditional. It tells you: given that A has occurred, what is the probability that it came from cause Bᵢ?'
Applications: medical testing ('Given a positive test, what is the probability you actually have the disease?'). Spam filtering. Machine learning.

5. Random Variables and Probability Distributions

Random Variable (X) : A function that assigns a real number to each outcome of a random experiment
Probability Distribution: The set of values X can take AND the probability of each value. ΣP(X=x) = 1.
Mean (Expected Value) : E(X) = Σ xᵢ pᵢ
Variance: Var(X) = E(X²) — [E(X)]²

6. Binomial Distribution

n INDEPENDENT trials. Each trial has TWO possible outcomes: SUCCESS (p) or FAILURE (q=1-p). p is CONSTANT across trials.
P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, for r = 0, 1, ..., n.
Mean = np. Variance = npq.

7. Exam Focus

Conditional probability. Multiplication theorem. Independent events.
Law of Total Probability. Bayes' Theorem — formula and application.
Random variable. Probability distribution. Mean and variance.
Binomial distribution — conditions, formula, mean (np), variance (npq).

8. Conclusion

Probability is the SCIENCE OF UNCERTAINTY:

CONDITIONAL: Information UPDATES probability. P(A) ≠ P(A|B).
BAYES: 'Given the evidence, what's the most likely cause?' The mathematical engine of learning from data.
BINOMIAL: Counting successes in repeated independent trials. 'The coin toss distribution — generalised to n tosses.'

'Bayes' Theorem is the mathematical form of rational belief: start with a prior, observe evidence, update to a posterior. It is the heart of statistics, machine learning, and scientific reasoning.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Conditional Probability and Multiplication Theorem

CONDITIONAL PROBABILITY: P(A|B) = P(A∩B) / P(B), provided P(B) > 0. Similarly P(B|A) = P(A∩B) / P(A). MULTIPLICATION THEOREM: P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B). INDEPENDENT EVENTS: A and B are INDEPENDENT if P(A∩B) = P(A)·P(B). Equivalently: P(A|B) = P(A) (occurrence of B doesn't change probability of A). NOTE: Independent ≠ Mutually Exclusive. If A and B are mutually exclusive (A∩B = ∅) and both have positive probability, they CANNOT be independent (since P(A∩B) = 0 ≠ P(A)·P(B)). PROPERTIES: P(A'|B) = 1 − P(A|B). For independent A, B: P(A∩B') = P(A)·P(B'). P(A'∩B') = P(A')·P(B').

MOST COMMON MISTAKE: confusing 'independent' with 'mutually exclusive.' Mutually exclusive events (A∩B=∅) are DEPENDENT (if A occurs, B cannot). Independent events can occur simultaneously. A coin flip and a die roll are independent; 'getting heads' and 'getting tails' on ONE flip are mutually exclusive.

Total Probability Theorem and Bayes' Theorem

PARTITION: Events E₁, E₂, ..., Eₙ form a PARTITION of S if: (i) Eᵢ ∩ Eⱼ = ∅ for i ≠ j, (ii) E₁ ∪ ... ∪ Eₙ = S, (iii) P(Eᵢ) > 0 for all i. TOTAL PROBABILITY THEOREM: For event A and partition {E₁,...,Eₙ}: P(A) = Σᵢ P(Eᵢ)·P(A|Eᵢ) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + ... + P(Eₙ)P(A|Eₙ). BAYES' THEOREM: P(Eₖ|A) = P(Eₖ)·P(A|Eₖ) / [Σᵢ P(Eᵢ)·P(A|Eᵢ)] = P(Eₖ)·P(A|Eₖ) / P(A). Prior probability: P(Eₖ). Posterior probability: P(Eₖ|A). Likelihood: P(A|Eₖ).

BAYES' THEOREM STRUCTURE: P(cause | effect) = P(cause)·P(effect | cause) / P(effect). You know the 'effect' (event A occurred) and want to find which 'cause' (E₁, E₂, etc.) is most likely. The denominator P(A) is always the TOTAL PROBABILITY (computed using the Total Probability Theorem). In exam problems: draw a tree diagram — branches from root are the Eᵢ, second branches are A|Eᵢ. Multiply along each path for the numerator products.

Random Variable and Probability Distribution

RANDOM VARIABLE X: a real-valued function defined on the sample space. PROBABILITY DISTRIBUTION: table showing all values x₁, x₂, ..., xₙ of X with corresponding probabilities p₁, p₂, ..., pₙ. Required: Σpᵢ = 1 and pᵢ ≥ 0. MEAN (Expected Value): E(X) = μ = Σxᵢpᵢ. VARIANCE: Var(X) = σ² = Σpᵢ(xᵢ−μ)² = E(X²) − [E(X)]² where E(X²) = Σxᵢ²pᵢ. STANDARD DEVIATION: σ = √Var(X). PROPERTIES: E(aX+b) = aE(X)+b. Var(aX+b) = a²Var(X).

For the CBSE probability distribution table: always verify that probabilities sum to 1. If a value k appears in the probabilities (like p₁ = k/6, p₂ = 2k/6, etc.), first find k using Σpᵢ = 1. The table itself is often the answer to the first part of the question; computing E(X) and Var(X) is the second part.

Binomial Distribution

CONDITIONS for Binomial: (1) Fixed number n of trials. (2) Each trial is independent. (3) Only two outcomes: success (probability p) and failure (probability q = 1−p). (4) p (and q) constant across trials. BINOMIAL DISTRIBUTION B(n, p): P(X = r) = ⁿCᵣ · pʳ · qⁿ⁻ʳ for r = 0, 1, 2, ..., n. MEAN: E(X) = np. VARIANCE: Var(X) = npq. STANDARD DEVIATION: σ = √(npq). MOST LIKELY VALUE (mode): r such that P(X=r) is maximum — often the integer part of (n+1)p.

BINOMIAL PROBABILITIES: P(X=0) = qⁿ (all failures). P(X=n) = pⁿ (all successes). P(X≥1) = 1 − P(X=0) = 1 − qⁿ. COMPLEMENT technique: P(at least one success) = 1 − P(no success) = 1 − qⁿ. This avoids summing many terms. Given mean np = 4 and variance npq = 2: q = 2/4 = 1/2, p = 1/2, n = 8. Mean and variance UNIQUELY determine n and p.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Treating mutually exclusive events as independent

✓ MUTUALLY EXCLUSIVE: P(A∩B) = 0. INDEPENDENT: P(A∩B) = P(A)·P(B). If A and B are mutually exclusive and both have positive probability, then P(A)·P(B) > 0 but P(A∩B) = 0 → they are DEPENDENT, NOT independent. The only case where mutually exclusive events are also independent is when at least one has probability 0. This distinction is a guaranteed exam MCQ.

WATCH OUT

✗ Writing Bayes' Theorem without computing the total probability denominator

✓ Bayes' theorem formula: P(Eₖ|A) = P(Eₖ)P(A|Eₖ) / P(A). The denominator P(A) MUST be computed using the Total Probability Theorem: P(A) = Σᵢ P(Eᵢ)P(A|Eᵢ). This is always a non-trivial calculation. Students often write the formula but then use only the specific Eₖ in the denominator (forgetting the other terms). Show the full denominator expansion explicitly.

WATCH OUT

✗ Computing Variance as E(X²) instead of E(X²) − [E(X)]²

✓ Variance = E(X²) − [E(X)]². Note: E(X²) ≠ [E(X)]². E(X²) = Σxᵢ²pᵢ (square x-values first, then weight). [E(X)]² = (Σxᵢpᵢ)². These are almost always different. The formula Var(X) = Σpᵢ(xᵢ−μ)² is equivalent but harder to compute. Use Var(X) = E(X²) − [E(X)]² as it's faster.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s 13.1–13.5

Exercises 13.1–13.5

Conditional probability; multiplication theorem; independence; Bayes' theorem with 2-3 events; probability distributions; mean and variance of distributions; binomial distribution applications

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· conditional-probability

A bag contains 5 red and 3 blue balls. Two balls are drawn one after another without replacement. Find the probability that both are red.

▶ Show solution

Let A = first ball is red, B = second ball is red. P(A) = 5/8. P(B|A) = 4/7 (4 red balls remain from 7 total after one red drawn). P(A∩B) = P(A)·P(B|A) = (5/8)·(4/7) = 20/56 = 5/14.

Q2MEDIUM· bayes-theorem

There are two bags. Bag I: 3 red, 4 black balls. Bag II: 5 red, 6 black balls. One bag is selected at random (equal probability) and one ball is drawn. The ball is red. Find the probability that it came from Bag I.

▶ Show solution

Let E₁ = Bag I selected, E₂ = Bag II selected, A = red ball drawn. P(E₁) = P(E₂) = 1/2. P(A|E₁) = 3/7. P(A|E₂) = 5/11. TOTAL PROBABILITY: P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) = (1/2)(3/7) + (1/2)(5/11) = 3/14 + 5/22 = 33/154 + 35/154 = 68/154 = 34/77. BAYES' THEOREM: P(E₁|A) = P(E₁)P(A|E₁) / P(A) = (1/2)(3/7) / (34/77) = (3/14) / (34/77) = (3/14) × (77/34) = 231/476 = 33/68.

Q3HARD· binomial-distribution

The probability that a student passes an exam is 2/3. In a group of 5 students, find: (i) P(at least 2 pass), (ii) the mean and variance of the number of students who pass.

▶ Show solution

X = number of students who pass ~ B(n=5, p=2/3, q=1/3). P(X=r) = ⁵Cᵣ·(2/3)ʳ·(1/3)^(5−r). (i) P(X ≥ 2) = 1 − P(X=0) − P(X=1). P(X=0) = ⁵C₀·(2/3)⁰·(1/3)⁵ = 1·1·1/243 = 1/243. P(X=1) = ⁵C₁·(2/3)¹·(1/3)⁴ = 5·(2/3)·(1/81) = 10/243. P(X ≥ 2) = 1 − 1/243 − 10/243 = 1 − 11/243 = 232/243. (ii) MEAN: E(X) = np = 5·(2/3) = 10/3. VARIANCE: Var(X) = npq = 5·(2/3)·(1/3) = 10/9. (Standard deviation = √(10/9) = √10/3.)

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•P(A|B) = P(A∩B)/P(B). P(A∩B) = P(A)·P(B|A). Independent: P(A∩B) = P(A)·P(B).
•Mutually exclusive ≠ independent. Mutually exclusive events with positive probability are DEPENDENT.
•Total Probability: P(A) = Σ P(Eᵢ)·P(A|Eᵢ) over a partition {E₁,...,Eₙ}.
•Bayes': P(Eₖ|A) = P(Eₖ)P(A|Eₖ) / [Σ P(Eᵢ)P(A|Eᵢ)]. Denominator = total probability.
•E(X) = Σxᵢpᵢ. Var(X) = E(X²) − [E(X)]². SD = √Var(X).
•Binomial B(n,p): P(X=r) = ⁿCᵣpʳqⁿ⁻ʳ. Mean = np. Variance = npq.
•P(X≥1) = 1 − qⁿ (complement method — always faster than summing individual terms).
•Σpᵢ = 1 for any probability distribution. Use this to find unknown k.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8-10 marks

Question type	Marks each	Typical count	What it tests
Short — Conditional Probability / Independence	2-3	1	P(A\|B) calculation; check independence; P(A∩B) using multiplication theorem
Medium — Bayes' Theorem	5-6	1	2-3 partition events; find posterior probability; real-world scenario (defective products, medical tests)
Short — Binomial Distribution	3-4	1	Find P(X=r) or P(X≥r); compute mean np and variance npq; identify n and p from given mean/variance

Prep strategy

Bayes' Theorem — always use a probability tree: (root) → E₁, E₂, E₃ with P(Eᵢ) labels → A|Eᵢ with P(A|Eᵢ) labels. Multiply along paths to get joint probabilities P(Eᵢ∩A). Sum for denominator P(A). Divide the specific path by the sum for P(Eᵢ|A). This tree method prevents denominator errors.
Binomial: verify conditions before applying (n fixed, independent, binary outcome, constant p). Write 'X ~ B(n, p)' to signal to examiners you've identified the distribution. State the formula P(X=r) = ⁿCᵣpʳqⁿ⁻ʳ explicitly before substituting.
For probability distribution tables: find k using Σpᵢ=1 first. Then compute E(X) = Σxpᵢ and E(X²) = Σx²pᵢ, then Var = E(X²) − [E(X)]². Show all four steps in sequence.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Medical Diagnostics and Bayes' Theorem

COVID-19 rapid antigen tests have a specificity (true negative rate) of ~95% and sensitivity (true positive rate) of ~80%. If 1% of the population is infected, Bayes' theorem computes the probability you actually have COVID given a positive test result: P(infected|positive) = P(+|infected)·P(infected) / P(+) ≈ (0.8×0.01) / (0.8×0.01 + 0.05×0.99) ≈ 0.008/0.0575 ≈ 14%. Despite a positive test, there's only a 14% chance of infection at 1% prevalence — counterintuitive, but mathematically exact via Bayes' theorem. Understanding this prevents both panic and false reassurance.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Bayes' Theorem: never skip writing the full denominator P(A) = Σ P(Eᵢ)P(A|Eᵢ) as a separate step. Examiners look for this total probability calculation before the final Bayes formula — it's typically 1-2 marks on its own.
Binomial distribution: state 'n = ..., p = ..., q = 1−p = ...' before computing any probabilities. Then write the formula P(X=r) = ⁿCᵣpʳqⁿ⁻ʳ and substitute. For 'at least' or 'at most' problems, always use the complement when r is small: P(X≥2) = 1 − P(X=0) − P(X=1) is faster than summing P(X=2)+...+P(X=n).

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Study the CENTRAL LIMIT THEOREM: for large n, the binomial distribution B(n,p) approximates a Normal distribution N(np, npq). This is why the bell curve appears everywhere in nature — averages of independent random variables converge to Normal regardless of the underlying distribution. It is the most important theorem in statistics
Explore MARKOV CHAINS: a sequence of random variables where P(Xₙ₊₁|Xₙ, Xₙ₋₁,...) = P(Xₙ₊₁|Xₙ) — the future depends only on the present, not the full history (Markov property). Bayes' theorem with conditional probability is the one-step version; Markov chains extend this to sequences. Google's PageRank, speech recognition, and weather forecasting all use Markov chain models

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (Probability)Very High

CUET (Mathematics)High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

TOTAL PROBABILITY THEOREM answers: 'Given the causes (E₁, E₂, ... with known prior probabilities), what is the probability of the effect (A)?' — P(A) = Σ P(Eᵢ)P(A|Eᵢ). BAYES' THEOREM answers: 'Given the effect (A) has occurred, what is the probability of cause Eₖ?' — P(Eₖ|A). Bayes' always uses the total probability as its denominator. In exam problems: if asked 'find P(A)' → Total Probability. If asked 'given A occurred, find P(Eₖ)' → Bayes. Most exam problems ask BOTH in parts (a) and (b).

VARIANCE Var(X) = E(X²) − [E(X)]² — a squared measure of spread, in square units. STANDARD DEVIATION σ = √Var(X) — the positive square root, in the same units as X. CBSE board asks for BOTH mean and variance for probability distribution questions. For Binomial distribution, Var(X) = npq is the standard result. Remember: standard deviation is √variance, not variance itself. If Var(X) = 10/9, then SD = √10/3.

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