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CBSE Class 12 Mathematics★ 6-8 marks · CBSE Class 12 Mathematics Chapter 13; Bayes' theorem and binomial distribution are high-yield for JEE

Probability

Probability is the MATHEMATICS of CHANCE. It tells us how LIKELY an event is to occur — from rolling dice to predicting the weather. This chapter builds on Class 11 concepts with CONDITIONAL probability, BAYES' theorem, random variables, and probability distributions. CBSE Class 12 Mathematics Chapter 13.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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

1Compute conditional probability and apply the multiplication theorem
2Test events for independence
3Apply the law of total probability and Bayes' theorem
4Build probability distributions and find mean and variance
5Use the binomial distribution for Bernoulli trials

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Why this chapter matters

Probability is the mathematics of uncertainty. Conditional probability, Bayes' theorem, random variables, and the binomial distribution are the logic behind risk, statistics, machine learning, and AI -- and reliable scoring topics in board and entrance exams.

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A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Mathematics: Probability

Sample space, events, and classical probability

Probability

'Probability does not tell us what WILL happen — it tells us how LIKELY each outcome IS.'

1. Chapter Overview

This chapter extends probability from Class 11 to more ADVANCED concepts. Topics include: CONDITIONAL PROBABILITY (the probability of event A given that B has occurred), the MULTIPLICATION THEOREM of probability, the LAW OF TOTAL PROBABILITY, BAYES' THEOREM (updating probabilities based on new evidence), RANDOM VARIABLES and their PROBABILITY DISTRIBUTIONS, the BINOMIAL DISTRIBUTION (Bernoulli trials), and the MEAN and VARIANCE of a random variable.

2. Conditional Probability

P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0.
'The probability of A GIVEN that B has already occurred.'
Key property: P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B).

Worked Example 1

Problem: A bag contains 4 red and 6 black balls. Two balls are drawn WITHOUT replacement. Find the probability that the second ball is red given that the first was red. Solution: P(A) = first ball is red = 4/10. P(A ∩ B) = P(both red) = (4/10)(3/9) = 12/90 = 2/15. Therefore P(B|A) = (2/15) / (4/10) = (2/15) × (10/4) = 20/60 = 1/3. Alternatively: after drawing one red, 3 red and 6 black remain ⇒ P(second red | first red) = 3/9 = 1/3. ✓

3. Multiplication Theorem and Independent Events

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) and P(B|A) = P(B).
'Independence means the occurrence of one event does NOT affect the probability of the other.'

4. Law of Total Probability

Let E₁, E₂, ..., Eₙ be a PARTITION of the sample space (mutually exclusive, exhaustive, non-empty).
For any event A: P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + ... + P(Eₙ)P(A|Eₙ).

5. Bayes' Theorem

P(Eᵢ|A) = [P(Eᵢ) × P(A|Eᵢ)] / Σⱼ[P(Eⱼ) × P(A|Eⱼ)]

'Bayes' theorem answers: given that A has happened, what is the probability that a SPECIFIC cause Eᵢ was responsible?'

Worked Example 2

Problem: There are two bags. Bag I contains 3 red and 4 black balls. Bag II contains 4 red and 5 black balls. One bag is chosen at random, and a ball is drawn from it. The ball is RED. Find the probability that it was drawn from Bag I. Solution: Let E₁ = Bag I chosen, E₂ = Bag II chosen. P(E₁) = P(E₂) = 1/2. P(Red|E₁) = 3/7. P(Red|E₂) = 4/9. By Bayes: P(E₁|Red) = [(1/2)(3/7)] / [(1/2)(3/7) + (1/2)(4/9)] = (3/7) / (3/7 + 4/9) = (3/7) / ((27+28)/63) = (3/7) / (55/63) = (3/7) × (63/55) = 27/55.

6. Random Variables and Probability Distributions

Random Variable

A RANDOM VARIABLE (RV) is a real-valued function defined on the sample space. 'It assigns a number to each outcome.'
Discrete RV: Takes COUNTABLE values. Example: number of heads in 3 tosses.

Probability Distribution

A table or function that assigns a probability to each value of the RV.
Conditions: Σ pᵢ = 1 and pᵢ ≥ 0 for all i.

Mean and Variance

Mean (Expected Value) : μ = E(X) = Σ xᵢ · pᵢ
Variance: Var(X) = σ² = Σ (xᵢ − μ)² · pᵢ = E(X²) − [E(X)]²
Standard deviation: σ = √Var(X)

7. Binomial Distribution

Bernoulli Trials

A trial with EXACTLY TWO outcomes: SUCCESS (S) or FAILURE (F).
Properties: (a) Fixed number of trials n. (b) Each trial is INDEPENDENT. (c) Probability of success p is CONSTANT.

Binomial Distribution

P(X = r) = ⁿCᵣ · pʳ · (1−p)ⁿ⁻ʳ, where r = 0, 1, 2, ..., n.
Notation: X ~ B(n, p).
Mean: μ = np. Variance: σ² = np(1−p). Standard deviation: σ = √(np(1−p)).

Worked Example 3

Problem: A die is thrown 6 times. If 'getting an even number' is a success, find the probability of exactly 4 successes. Solution: p = P(even) = 3/6 = 1/2. n = 6. r = 4. P(X=4) = ⁶C₄ × (1/2)⁴ × (1/2)² = 15 × (1/16) × (1/4) = 15/64.

8. Comparison Table: Classical vs Conditional vs Binomial Probability

Type	Formula	When to Use
Classical	P(E) = n(E)/n(S)	All outcomes equally likely
Conditional	P(A	B) = P(A∩B)/P(B)
Bayes' Theorem	P(Eᵢ	A) = P(Eᵢ)P(A
Binomial	P(X=r) = ⁿCᵣ pʳ(1−p)ⁿ⁻ʳ	Fixed n independent trials, 2 outcomes each

9. Common Mistakes

Confusing P(A|B) with P(B|A): They are NOT the same in general. P(A|B) is 'probability of A given B'. Bayes' theorem relates the two.
Assuming independence without checking: Just because two events seem unrelated does not mean they are independent. ALWAYS verify: P(A∩B) = P(A)P(B)?
Forgetting to condition properly: 'What is the probability that a randomly selected student is a girl who studies science?' vs 'What is the probability that a science student is a girl?' — these are VERY different questions.
Binomial conditions: For binomial, EVERY trial must be independent with CONSTANT probability of success. Drawing without replacement is NOT binomial.

10. CBSE Exam Focus

Conditional probability — definition and computation
Multiplication theorem and independent events
Bayes' theorem — especially word problems with 'cause → effect' structure
Probability distribution of a random variable — finding missing probabilities, computing mean and variance
Binomial distribution — ⁿCᵣ pʳ(1−p)ⁿ⁻ʳ, mean = np, variance = np(1−p)

11. Self-Test

Q1: A pair of dice is thrown. Find the probability that the sum is 7 given that the first die shows a 3. A1: First die = 3. Favourable outcomes for sum 7: (3,4). P = 1/6.

Q2: A coin is tossed 5 times. Find the probability of getting at least 3 heads. A2: Binomial with n=5, p=1/2. P(X≥3) = P(3)+P(4)+P(5) = ⁵C₃(1/2)⁵ + ⁵C₄(1/2)⁵ + ⁵C₅(1/2)⁵ = (10+5+1)/32 = 16/32 = 1/2.

Q3: A random variable X has the distribution: X=0 (p=0.1), X=1 (p=0.3), X=2 (p=0.4), X=3 (p=k). Find k, E(X), and Var(X). A3: Σp = 1 ⇒ 0.1+0.3+0.4+k = 1 ⇒ k = 0.2. E(X) = 0(0.1)+1(0.3)+2(0.4)+3(0.2) = 0+0.3+0.8+0.6 = 1.7. E(X²) = 0(0.1)+1(0.3)+4(0.4)+9(0.2) = 0+0.3+1.6+1.8 = 3.7. Var(X) = 3.7 − (1.7)² = 3.7 − 2.89 = 0.81.

Q4: A bag contains 5 white and 3 black balls. Two balls are drawn. Find the probability that they are of different colours. A4: P(different) = P(white then black) + P(black then white) = (5/8)(3/7) + (3/8)(5/7) = 15/56 + 15/56 = 30/56 = 15/28.

Q5: In a test, an examinee either guesses or copies or knows the answer with probabilities 0.1, 0.2, and 0.7 respectively. The probability that he answers correctly is 0.2 if he guesses, 0.8 if he copies. Find the probability that he knew the answer given that he answered correctly. A5: Let G=guess, C=copy, K=know, R=correct. P(G)=0.1, P(C)=0.2, P(K)=0.7. P(R|G)=0.2, P(R|C)=0.8, P(R|K)=1. P(R) = 0.1(0.2)+0.2(0.8)+0.7(1) = 0.02+0.16+0.7 = 0.88. By Bayes: P(K|R) = 0.7(1)/0.88 = 70/88 = 35/44.

12. Conclusion

Probability is the MATHEMATICAL FRAMEWORK for dealing with UNCERTAINTY:

CONDITIONAL: 'What is the probability of A given B?' — the most important concept in probability.
BAYES: 'How to update beliefs when new evidence arrives.' — the foundation of machine learning and AI.
BINOMIAL: 'The distribution of count of successes in n independent trials.' — the simplest and most important discrete distribution.
RANDOM VARIABLES: 'Assigning numbers to outcomes — the bridge between probability and statistics.'

'Probability is not just about dice and cards — it is the logic of uncertainty, essential for science, medicine, finance, and AI.'

Key formulas & results

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

Conditional probability

P(A|B) = P(A and B)/P(B)

Multiplication theorem: P(A and B) = P(A)P(B|A).

Bayes' theorem

P(Ei|A) = P(Ei)P(A|Ei) / sum_j P(Ej)P(A|Ej)

Updates the probability of a cause given an effect.

Mean and variance

E(X) = sum xi pi; Var(X) = E(X^2) - [E(X)]^2

For a discrete random variable.

Binomial distribution

P(X=r) = nCr p^r (1-p)^(n-r); mean = np, variance = np(1-p)

For n independent Bernoulli trials.

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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

✗ Confusing P(A|B) with P(B|A)

✓ These are generally different; Bayes' theorem is what relates the two.

WATCH OUT

✗ Assuming events are independent without checking

✓ Verify independence with P(A and B) = P(A)P(B); unrelated-seeming events are not automatically independent.

WATCH OUT

✗ Applying the binomial distribution to drawing without replacement

✓ Binomial requires constant success probability and independent trials; sampling without replacement violates this.

WATCH OUT

✗ Forgetting the probabilities must sum to 1

✓ In a probability distribution, sum of all pi = 1 and each pi >= 0; use this to find missing values.

NCERT exercises (with solutions)

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

Conditional probability and independence

Conditional probability, multiplication theorem, independence

Questions

Bayes' theorem

Total probability and cause-effect word problems

Questions

Distributions

Random variables, mean, variance, binomial distribution

Questions

Practice problems

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

Q1EASY· Conditional

A pair of dice is thrown. Find the probability the sum is 7 given the first die shows 3.

▶ Show solution

With the first die fixed at 3, the only way to get sum 7 is (3,4). So P = 1/6.

Q2MEDIUM· Binomial

A coin is tossed 5 times. Find the probability of at least 3 heads.

▶ Show solution

n = 5, p = 1/2. P(X >= 3) = (5C3 + 5C4 + 5C5)/32 = (10 + 5 + 1)/32 = 16/32 = 1/2.

Q3MEDIUM· Distribution

X takes values 0,1,2,3 with probabilities 0.1, 0.3, 0.4, k. Find k, E(X), and Var(X).

▶ Show solution

Sum = 1 gives k = 0.2. E(X) = 0.3 + 0.8 + 0.6 = 1.7. E(X^2) = 0.3 + 1.6 + 1.8 = 3.7. Var(X) = 3.7 - 1.7^2 = 0.81.

Q4MEDIUM· Probability

A bag has 5 white and 3 black balls. Two are drawn. Find the probability they are different colours.

▶ Show solution

P = (5/8)(3/7) + (3/8)(5/7) = 15/56 + 15/56 = 30/56 = 15/28.

Q5HARD· Bayes

An examinee guesses, copies, or knows an answer with probabilities 0.1, 0.2, 0.7; correct with probability 0.2, 0.8, 1 respectively. Find P(knew | correct).

▶ Show solution

P(correct) = 0.1(0.2) + 0.2(0.8) + 0.7(1) = 0.88. P(knew | correct) = 0.7/0.88 = 35/44.

5-minute revision

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

•Conditional probability P(A|B) = P(A and B)/P(B).
•Independent events: P(A and B) = P(A)P(B).
•Law of total probability sums over a partition of the sample space.
•Bayes' theorem updates the probability of a cause given an effect.
•Probability distribution: sum pi = 1, pi >= 0.
•Mean E(X) = sum xi pi; Var(X) = E(X^2) - [E(X)]^2.
•Binomial: P(X=r) = nCr p^r (1-p)^(n-r); mean np, variance np(1-p).

CBSE marks blueprint

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

Typical chapter weightage: 6-8 marks across the chapter

Question type	Marks each	Typical count	What it tests
Bayes' theorem	4	1	Cause-effect word problems
Binomial / distribution	3	1	Binomial probabilities, mean, variance
Conditional / independence	2-3	1	Conditional probability and independence

Prep strategy

Distinguish P(A|B) from P(B|A) carefully
Set up Bayes problems with a clear partition
Check binomial conditions before using the formula
Use sum of probabilities = 1 for distributions

Where this shows up in the real world

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

Medical diagnosis

Bayes' theorem updates the probability of disease from test results.

Machine learning and AI

Bayesian reasoning underlies spam filters, classifiers, and probabilistic models.

Quality control

The binomial distribution models defect counts in manufacturing samples.

Exam strategy

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

Write the partition clearly in Bayes problems
Keep P(A|B) and P(B|A) distinct
Verify binomial conditions before applying the formula
Use sum of probabilities = 1 to find unknowns

Going beyond the textbook

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

Explore the Poisson distribution as a limit of the binomial.
Analyse conditional expectation and the law of total expectation.

Where else this chapter is tested

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

CBSE Class 12 Mathematics examHigh

JEE Main (Probability)High

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Questions students ask

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

Bayes' theorem lets us reverse a conditional probability. Often we know how likely an effect is given each possible cause (for example, the chance of testing positive given a disease), but we really want the probability of a cause given the observed effect (the chance of having the disease given a positive test). Bayes' theorem combines the prior probabilities of the causes with these conditional probabilities to update our belief about which cause is responsible, which is why it is fundamental to diagnosis, machine learning, and decision-making under uncertainty.

The binomial distribution applies to a fixed number n of independent trials, each with exactly two outcomes (success or failure) and the same probability p of success on every trial -- these are called Bernoulli trials. For example, tossing a coin a fixed number of times fits this model. Drawing balls without replacement does not, because removing a ball changes the composition and so the success probability is no longer constant and the trials are not independent.

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