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

Probability Class 12: notes and important questions

Revision notes, 39 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.

  • 39 questions
  • 12 multiple choice, 3 assertion–reason, 8 very short answer, 8 short answer, 5 long answer, 3 case study
  • About 7 hours to master

Unit VI Probability: 8 marks of the 80-mark paper — CBSE Curriculum 2025-26 Mathematics (041), https://cbseacademic.nic.in/web_material/CurriculumMain26/SrSec/Maths_SrSec_2025-26.pdf.

Revision notes

Probability — revision notes

1. Conditional probability and multiplication theorem

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

2. Independent events

  • \(A, B\) independent \(\iff P(A \cap B) = P(A)P(B)\) \(\iff P(A \mid B) = P(A)\).
  • If \(A, B\) are independent, so are \(A\prime\) and \(B\), \(A\) and \(B\prime\), \(A\prime\) and \(B\prime\). \(P(\text{at least one}) = 1 - P(A\prime)P(B\prime)\cdots\)
  • 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: \(P(A) = \sum P(E_i)P(A \mid E_i)\).
  • Bayes: \(P(E_i \mid A) = \dfrac{P(E_i)P(A \mid E_i)}{\sum_j P(E_j)P(A \mid E_j)}\). 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 \(p_i \ge 0\), \(\sum p_i = 1\).
  • Mean (expectation) \(E(X) = \sum x_ip_i\). 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 \(\dfrac{4}{12} = \dfrac13\).

Worked example 2

Machines M1, M2 make \(60\%\) and \(40\%\) of bolts, with \(3\%\) and \(5\%\) defective. \(P(D) = 0.018 + 0.020 = 0.038\); \(P(M_2 \mid D) = \dfrac{0.020}{0.038} = \dfrac{10}{19}\).

Worked example 3

Two coins are tossed; \(X\) = number of heads: \(0, 1, 2\) with \(\tfrac14, \tfrac12, \tfrac14\); \(E(X) = 1\).

Common errors

  • Dividing by the wrong event: in \(P(A \mid B)\) 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.

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. 16 of the 39 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 \(P(A\prime \mid B)\) equals

  1. (a)\(0.2\)
  2. (b)\(0.4\)
  3. (c)\(0.5\)
  4. (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 \(P(X \ge 2)\) equals

  1. (a)\(\tfrac{7}{10}\)
  2. (b)\(\tfrac{3}{10}\)
  3. (c)\(\tfrac12\)
  4. (d)\(\tfrac{4}{10}\)
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

  1. (a)\(1\)
  2. (b)\(0.3\)
  3. (c)\(0.7\)
  4. (d)\(\tfrac13\)

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).

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).