Class 12 · Chapter 13 · Probability unit (8 of 80 marks)

Probability Class 12: notes and important questions

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 56 board-style questions.

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56 questions
15 multiple choice, 3 assertion–reason, 13 very short answer, 15 short answer, 6 long answer, 4 case study
About 7 hours to master

Unit VI Probability: 8 marks of the 80-mark paper — CBSE Curriculum 2025-26 Mathematics (041) (official PDF).

NCERT textbook: Chapter 13 on ncert.nic.in (PDF)

Revision notes

Probability — revision notes

1. Conditional probability and multiplication theorem

$P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ , $P (B) \neq 0$ . It is a probability: $P (A' ∣ B) = 1 - P (A ∣ B)$ .
$P (A \cap B) = P (A) P (B ∣ A) = P (B) P (A ∣ B)$ ; for three events $P (A \cap B \cap C) = P (A) P (B ∣ A) P (C ∣ A \cap B)$ (draws without replacement).
For equally likely outcomes, $P (A ∣ B)$ = (outcomes in $A \cap B$ ) ÷ (outcomes in $B$ ) — the given event becomes the new sample space.

2. Independent events

$A, B$ independent $⟺ P (A \cap B) = P (A) P (B)$ $⟺ P (A ∣ B) = P (A)$ .
If $A, B$ are independent, so are $A'$ and $B$ , $A$ and $B'$ , $A'$ and $B'$ . $P (at least one) = 1 - P (A') P (B') \dots$
Mutually exclusive ( $A \cap B = \emptyset$ ) is not the same as independent; with non-zero probabilities they cannot be both.

3. Total probability and Bayes’ theorem

If $E_{1}, \dots, E_{n}$ partition the sample space: $P (A) = \sum P (E_{i}) P (A ∣ E_{i})$ .
Bayes: $P (E_{i} ∣ A) = \frac{P (E_{i}) P (A ∣ E_{i})}{\sum_{j} P (E_{j}) P (A ∣ E_{j})}$ . Name the events first; a tree helps.

4. Random variable and its mean

A random variable assigns a real number to each outcome. Its probability distribution lists the values $x_{i}$ with $p_{i} \geq 0$ , $\sum p_{i} = 1$ .
Mean (expectation) $E (X) = \sum x_{i} p_{i}$ . A game is fair when the expected net gain is $0$ .

Worked example 1

A card is drawn from a pack of 52. Given it is a face card (J, Q, K), the probability that it is a king is $\frac{4}{12} = \frac{1}{3}$ .

Worked example 2

Machines M1, M2 make $60 %$ and $40 %$ of bolts, with $3 %$ and $5 %$ defective. $P (D) = 0.018 + 0.020 = 0.038$ ; $P (M_{2} ∣ D) = \frac{0.020}{0.038} = \frac{10}{19}$ .

Worked example 3

Two coins are tossed; $X$ = number of heads: $0, 1, 2$ with $\frac{1}{4}, \frac{1}{2}, \frac{1}{4}$ ; $E (X) = 1$ .

Common errors

Dividing by the wrong event: in $P (A ∣ B)$ the given event $B$ goes in the denominator.
Assuming independence without justification, or confusing it with mutually exclusive.
In Bayes’ questions, forgetting a branch in the denominator or not defining $E_{1}, E_{2}, \dots$ and $A$ .
Probability distributions whose probabilities do not add to $1$ — always check.
Treating “without replacement” draws as independent.

Board-exam tips

In Bayes’ questions, marks are given for defining events, writing the prior and likelihood values, the formula and the final answer: show each line.
The binomial distribution is not in the current syllabus; only the mean of a random variable is asked (no variance).

Topics in this chapter: Conditional probability · Independent events · Random variable and its probability distribution · Mean of a random variable · Theorem of total probability · Multiplication theorem on probability · Bayes' theorem.

Formulas for this chapter: Chapter 13 on the Class 12 formula sheet (free PDF and A3 poster).

Probability Class 12: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 6 pages · 320 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choiceIndependent events and multiplication theorem

Independent events $A$ and $B$ have $P (A) = \frac{1}{4}$ and $P (B) = \frac{2}{5}$ . What is $P (A \cup B)$ ?

Show worked answer

$\begin{aligned} P (A \cap B) & = \frac{1}{4} \times \frac{2}{5} \\ = \frac{1}{10} \end{aligned}$ , so $\begin{aligned} P (A \cup B) & = \frac{5}{20} + \frac{8}{20} - \frac{2}{20} \\ = \frac{11}{20} \end{aligned}$ A1
Correct option: (b) $\frac{11}{20}$

Q2·1 mark·Multiple choiceConditional probability

$P (B) = \frac{2}{5}$ and $P (A \cap B) = \frac{1}{5}$ . Then $P (A ∣ B)$ is

Q3·1 mark·Assertion–reasonIndependent events and multiplication theorem

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): If $A$ and $B$ are independent events with $P (A) = P (B) = \frac{1}{2}$ , then $A$ and $B$ are mutually exclusive.

Reason (R): For independent events, $P (A \cap B) = P (A) P (B)$ .

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerConditional probability

A fair die is thrown once. $E$ is the event 'the number is even' and $F$ is 'the number is greater than $3$ '. Find $P (E ∣ F)$ .

Show worked answer

Answer:

\frac{2}{3}

$F = {4, 5, 6}$ and $E \cap F = {4, 6}$ M1
$\begin{aligned} P (E ∣ F) & = \frac{2 / 6}{3 / 6} \\ = \frac{2}{3} \end{aligned}$ A1

Q5·2 marks·Short answerIndependent events and multiplication theorem

A bag holds $3$ red and $5$ black balls. Two balls are drawn one after the other without replacement. Find the probability that both are red.

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerRandom variables and their mean

A fair coin is tossed twice and $X$ is the number of tails. Find the probability distribution of $X$ and its mean.

Show worked answer

Answer:

\frac{1}{4}, \frac{1}{2}, \frac{1}{4}

; mean

1

$X$ takes the values $0, 1, 2$ M1
$P (X = 0) = \frac{1}{4}$ , $P (X = 1) = \frac{1}{2}$ , $P (X = 2) = \frac{1}{4}$ A1
Mean $= 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 2 \times \frac{1}{4} = 1$ A1

Q7·3 marks·Short answerIndependent events and multiplication theorem

The probability that Arjun hits a target is $\frac{1}{3}$ and that Bela hits it is $\frac{1}{4}$ . They shoot independently. Find the probability that (a) the target is hit, (b) exactly one of them hits it.

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerTotal probability and Bayes' theorem

Three couriers A, B and C deliver $50 %$ , $30 %$ and $20 %$ of a shop's parcels. Of their parcels, $2 %$ , $3 %$ and $4 %$ respectively arrive late. A parcel is found to be late. Find the probability that it was delivered by A, and the probability that it was delivered by C.

Tree diagram: courier A 50%, B 30%, C 20%. From A, late 2% and on time 98%; from B, late 3% and on time 97%; from C, late 4% and on time 96%. — Couriers A, B, C and late parcels

Show worked answer

Answer:

\frac{10}{27}

and

\frac{8}{27}

$P (A) = \frac{1}{2}$ , $P (B) = \frac{3}{10}$ , $P (C) = \frac{1}{5}$ ; $P (L ∣ A) = \frac{2}{100}$ , $P (L ∣ B) = \frac{3}{100}$ , $P (L ∣ C) = \frac{4}{100}$ M1
Total probability: $P (L) = \frac{1}{100} + \frac{9}{1000} + \frac{8}{1000}$ M1
$= \frac{27}{1000}$ A1
Bayes' theorem: $\begin{aligned} P (A ∣ L) & = \frac{10 / 1000}{27 / 1000} \\ = \frac{10}{27} \end{aligned}$ A1
$\begin{aligned} P (C ∣ L) & = \frac{8 / 1000}{27 / 1000} \\ = \frac{8}{27} \end{aligned}$ A1

Q9·5 marks·Long answerRandom variables and their mean

A bag holds $4$ red and $2$ white balls. Two balls are drawn at random without replacement, and $X$ is the number of white balls drawn. Find the probability distribution of $X$ and the mean of $X$ .

Section E style · Case study (4 marks)

Q10·4 marks·Case studyTotal probability and Bayes' theorem

Screening test. A screening test finds a condition in $9$ out of every $10$ people who have it, but also wrongly flags $1$ in every $20$ people who do not. One person in $10$ in a town has the condition. A person is tested at random.

Find the probability that the person has the condition and tests positive. [1 mark]
Find the probability that the person tests positive. [1 mark]
A person tests positive. Find the probability that they have the condition. [2 marks]

Tree diagram: has the condition 1/10, does not 9/10. If they have it: positive 9/10, negative 1/10. If they do not: positive 1/20, negative 19/20. — Screening test: condition, then test result

Show worked answer

$\frac{1}{10} \times \frac{9}{10} = \frac{9}{100}$ A1
$\begin{aligned} \frac{9}{100} + \frac{9}{10} \times \frac{1}{20} & = \frac{9}{100} + \frac{9}{200} \\ = \frac{27}{200} \end{aligned}$ A1
Bayes' theorem: $\frac{9 / 100}{27 / 200}$ M1
$= \frac{18}{27} = \frac{2}{3}$ A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 12 assertion–reason questions · the Class 12 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

Compute conditional probabilities from counts and formulas, use the multiplication theorem and find k in a probability distribution.

Read first: 1. Conditional probability and multiplication theorem; 4. Random variable and its mean 8 practice questions · checkpoint: 4 questions, 6 marks, pass 80%

Practise step 1

Step 2

Board standard

Test independence, apply total probability and Bayes’ theorem, and find the mean of a random variable at board 2- and 3-mark level.

Read first: 2. Independent events; 3. Total probability and Bayes’ theorem 20 practice questions · checkpoint: 4 questions, 9 marks, pass 75%

Practise step 2

Step 3

Full marks on long answers

Write complete 5-mark Bayes and random-variable answers (events defined, tree/table, formula, conclusion) and handle probability case studies.

Read first: 3. Total probability and Bayes’ theorem; Worked example 2; Board-exam tips 5 practice questions · checkpoint: 3 questions, 14 marks, pass 70%

Practise step 3

Step 4

95+ stretch (HOTS)

Handle proofs about conditional probability and independence, sequential evidence in Bayes, and distributions built from without-replacement experiments.

Read first: 2. Independent events; 3. Bayes’ theorem; 4. Random variable 6 practice questions · checkpoint: 3 questions, 13 marks, pass 60%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 21 of the 56 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceConditional probability

If $P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.2$ , then $P (A' ∣ B)$ equals

(a) $0.2$
(b) $0.4$
(c) $0.5$
(d) $0.8$

Q2·1 mark·Multiple choiceRandom variable and its probability distribution

A random variable $X$ has $P (X = x) = k (x + 1)$ for $x = 0, 1, 2, 3$ (and $0$ otherwise). Then $P (X \geq 2)$ equals

(a) $\frac{7}{10}$
(b) $\frac{3}{10}$
(c) $\frac{1}{2}$
(d) $\frac{4}{10}$

Q3·1 mark·Multiple choiceMean of a random variable

A random variable $X$ takes the values $- 1$ , $0$ and $2$ with probabilities $0.3$ , $0.2$ and $0.5$ . The mean of $X$ is

(a) $1$
(b) $0.3$
(c) $0.7$
(d) $\frac{1}{3}$

Worked solution·2 marks·Very short answerMean of a random variable

A fair die is thrown twice and $X$ is the number of sixes. Find the mean of $X$ .

Full worked solution Free, no sign-in

$P (X = 0) = \frac{25}{36}$ , $P (X = 1) = \frac{10}{36}$ , $P (X = 2) = \frac{1}{36}$ A1
$E (X) = \frac{10}{36} + \frac{2}{36} = \frac{1}{3}$ A1

Every question here has a full worked solution. See them all →

Stretch yourself: original probability problems at Senior level (ages 16 to 18), with hints and full solutions: Problem S101 · Problem S29 · Problem S102 · Problem S30. All 20 →

Knowledge organiser: Probability on one page (key facts, formulas and mistakes to avoid; printable PDF)

Formulas in this topic: Bayes' theorem. All formulas A–Z

Where marks are lost in Probability

Events not defined in a Bayes question, so the method marks are lost. Fix: start with "Let E₁: …, E₂: …, A: …" and list P(Eᵢ) and P(A|Eᵢ).
Denominator of Bayes missing a branch. Fix: compute P(A) by total probability first, on its own line, then divide.
Using P(A ∩ B) = P(A)P(B) without being told the events are independent. Fix: check independence or use P(A)P(B|A).
Confusing mutually exclusive with independent. Fix: mutually exclusive means P(A ∩ B) = 0; independent means P(A ∩ B) = P(A)P(B).
A probability distribution that does not sum to 1 (often from a missed value of X). Fix: list every possible value of X and add the probabilities before finding the mean.
Treating draws without replacement as independent. Fix: change the denominators after each draw (multiplication theorem).
Answering a "fair game" question without the sign of E(X). Fix: state the mean net gain and interpret it (positive favours the player).

Exam strategy: how answers are step-marked · competency-based questions

Probability in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 18, 20, 25, 36
Sample paper 2 (Board standard): questions 20, 31, 38
Sample paper 3 (Above board): questions 18, 20, 25, 36
Sample paper 4 (Above board): questions 20, 31, 38
Sample paper 5 (95+ challenge): questions 18, 20, 25, 36

Want it against the clock? Take a timed 30-mark chapter test on Probability, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Preparing for JEE or CUET? Try the JEE and CUET-style questions on statistics and probability.

Probability: common questions

What is in CBSE Class 12 Maths Chapter 13, Probability?

The topics are: Conditional probability, Independent events, Random variable and its probability distribution, Mean of a random variable, Theorem of total probability, Multiplication theorem on probability, Bayes' theorem. This page has revision notes for each, and 56 board-style practice questions with step-marked solutions.

How many marks is Probability worth?

It is in the Probability unit, which carries 8 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Probability?

Events not defined in a Bayes question, so the method marks are lost. Fix: start with "Let E₁: …, E₂: …, A: …" and list P(Eᵢ) and P(A|Eᵢ). Denominator of Bayes missing a branch. Fix: compute P(A) by total probability first, on its own line, then divide.

Are the Probability important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

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Revision notes

Probability — revision notes

1. Conditional probability and multiplication theorem

2. Independent events

3. Total probability and Bayes’ theorem

4. Random variable and its mean

Worked example 1

Worked example 2

Worked example 3

Common errors

Board-exam tips

Probability Class 12: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Probability

Probability in our sample papers

Probability: common questions

Work through the Route to 95 with your progress saved