If \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \cap B) = 0.2\), then \(P(A\prime \mid B)\) equals
- (a)\(0.2\)
- (b)\(0.4\)
- (c)\(0.5\)
- (d)\(0.8\)
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.
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.
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.
Step 4
95+ stretch (HOTS)
Handle proofs about conditional probability and independence, sequential evidence in Bayes, and distributions built from without-replacement experiments.
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.
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
- (a)\(\tfrac{7}{10}\)
- (b)\(\tfrac{3}{10}\)
- (c)\(\tfrac12\)
- (d)\(\tfrac{4}{10}\)
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)\(\tfrac13\)
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 →
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 Essentials or the free trial).