Probability

2.1 Core concepts

• Probability is treated as a function on outcomes (axiomatic approach of Kolmogorov), with discrete sample spaces and the addition rule already known from Class XI (NCERT §13.1, p. 406). The non-negativity, normalisation (P(S) = 1) and σ-additivity axioms are assumed throughout.
• Conditional probability of E given F (with P(F) ≠ 0) is defined as P(E|F) = P(E ∩ F)/P(F); intuitively, the occurrence of F reduces the sample space from S to F and we measure E inside F (NCERT §13.2, p. 408, Definition 1). The denominator P(F) "renormalises" so that P(F|F) = 1.
• Three properties of conditional probability are stated and proved: (i) P(S|F) = P(F|F) = 1, (ii) P((A ∪ B)|F) = P(A|F) + P(B|F) − P((A ∩ B)|F) with the disjoint-case corollary P((A ∪ B)|F) = P(A|F) + P(B|F), and (iii) P(E′|F) = 1 − P(E|F) (NCERT §13.2.1, pp. 408-409). All three reduce to the usual Kolmogorov properties applied to the restricted sample space F.
• Worked tossing/dice examples (Examples 1-7) show that the conditional definition applies even when elementary outcomes are not equally likely; the tree-diagram example with one coin then die illustrates P(E|F) for unequal-probability sample points (NCERT §13.2, pp. 409-413).
• The multiplication theorem on probability follows directly from the definition: P(E ∩ F) = P(E)·P(F|E) = P(F)·P(E|F), provided both P(E) ≠ 0 and P(F) ≠ 0 (NCERT §13.3, p. 415). It is the basic tool for "without replacement" tree problems.
• For three events the rule extends to P(E ∩ F ∩ G) = P(E)·P(F|E)·P(G|EF), and the urn/card examples (Examples 8-9) apply it to drawing without replacement (NCERT §13.3, pp. 415-417). The general n-event chain rule has n factors.
• Independent events are defined by P(F|E) = P(F) and P(E|F) = P(E); equivalently, P(E ∩ F) = P(E)·P(F). Three events A, B, C are mutually independent iff all pairwise products and the triple product hold (NCERT §13.4, pp. 417-418, Definitions 2-3 + Remark iv). Pairwise independence does NOT imply mutual independence.
• Independence is sharply distinguished from mutual exclusivity: "independent" is about probabilities, "mutually exclusive" is about sets; two events with nonzero probability cannot be both (NCERT §13.4, p. 418, Remark ii). This Remark is paraphrased on virtually every CUET paper.
• A partition {E₁, E₂, …, Eₙ} of S consists of pairwise disjoint, exhaustive events each of nonzero probability (NCERT §13.5.1, p. 423). The partition is the conceptual setting for both the total-probability and Bayes formulas.
• Theorem of total probability: for any event A, P(A) = Σⱼ P(Eⱼ)·P(A|Eⱼ); proved by writing A = (A ∩ E₁) ∪ … ∪ (A ∩ Eₙ) as a disjoint union and applying the multiplication rule (NCERT §13.5.2, p. 424). The formula is the "forward direction" computation: given hypotheses, find the overall probability of A.
• Bayes' theorem: for a partition {E₁, …, Eₙ} and any event A with P(A) ≠ 0, P(Eᵢ|A) = P(Eᵢ)·P(A|Eᵢ) / Σⱼ P(Eⱼ)·P(A|Eⱼ); the Eᵢ are called hypotheses, P(Eᵢ) the priori probabilities and P(Eᵢ|A) the posteriori probabilities (NCERT §13.5, pp. 425-426). The denominator is exactly the total-probability formula for P(A); Bayes inverts the conditioning.
• Applications worked in full: drawing a red ball from one of two bags (Ex. 16), the two-coins-in-a-box problem (Ex. 17), HIV reliability (Ex. 18), defective bolts from three machines (Ex. 19), late-doctor transport (Ex. 20), and the truth-teller die (Ex. 21) (NCERT pp. 426-430).
• A random variable is a real-valued function on the sample space, illustrated by counting heads in two tosses (NCERT p. 430). Expectation and variance of a discrete random variable are touched on only in passing; full development is reserved for Class XII miscellaneous exercises and beyond.
• Conditional-probability intuition: imagine the sample space "shrinking" from S to F once you learn F has occurred. The conditional probability P(E|F) measures the share of F that is also in E. When F = S (no extra information), P(E|S) = P(E ∩ S)/P(S) = P(E), so conditional probability generalises unconditional probability.
• A second, equivalent definition of independence: E and F are independent iff knowing F has occurred does not change the probability of E, i.e. P(E|F) = P(E). This makes the connection between the "probability" formulation (P(E ∩ F) = P(E) P(F)) and the "information" interpretation explicit (NCERT §13.4, p. 418).
• Bayes' theorem can be re-read as a belief update: the prior P(Eᵢ) is updated by multiplying by the likelihood ratio P(A|Eᵢ) / P(A), giving the posterior P(Eᵢ|A). This is the cornerstone of statistical inference.
• The "law of total probability" reading: P(A) is a weighted average of the conditional probabilities P(A|Eⱼ), weighted by the priors P(Eⱼ). When all P(A|Eⱼ) are equal, P(A) equals that common value, regardless of the priors.
• Independence of a class of events is a strong condition. For n events to be mutually independent, all 2ⁿ − n − 1 joint-probability relations must hold (every subset of size ≥ 2). For three events that means four conditions (three pairwise + one triple). NCERT only requires verifying these four for Class XII.
• The complement rule for independence: if E, F are independent, so are E and F′, and E′ and F′. This often shortens calculations: rather than computing P(at least one) directly, use 1 − P(none).
• A classic Bayes' set-up not in NCERT but mirrored in Exercise 13.3: a disease has 1% prevalence; a test has 99% sensitivity and 95% specificity. Then P(disease | positive) ≈ 0.167, not 0.99 — illustrating how rarity of the disease moderates the posterior, a frequent CUET-style insight.
• Random-variable terminology: a random variable X assigns a real number to each outcome of the sample space. Every later application (binomial distribution, mean and variance computations) builds on this definition.
• The two-toss example: tossing a coin twice gives S = {HH, HT, TH, TT}; let X count heads. Then X(HH) = 2, X(HT) = X(TH) = 1, X(TT) = 0, so X takes values 0, 1, 2 with probabilities 1/4, 1/2, 1/4 — the simplest non-trivial random variable, used throughout statistical inference.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 13.1 / 13.2 (p. 412): Tree diagram for the experiment "toss a coin; if head, toss again; if tail, throw a die" — fixes the technique of multiplying along branches to get elementary-event probabilities (1/4, 1/4, 1/12 each for the six die outcomes). Tree diagrams are the universal solvent for multi-stage probability problems.
• Fig 13.3 (p. 420): Venn diagram showing E = (E ∩ F) ∪ (E ∩ F′) used to prove that if E and F are independent then so are E and F′.
• Fig 13.4 (p. 424): Partition diagram of S into E₁, E₂, …, Eₙ with A cut by each Eᵢ — the visual basis of the total-probability proof. Sketching this figure is often the fastest way to set up a Bayes problem.
• Bolt-machine and bag-ball Bayes set-ups (pp. 426-429): the standard table format — list P(Eᵢ), then P(A|Eᵢ), then plug into Bayes — is the template for almost every Bayes MCQ.
• Process — conditional probability: (i) identify the conditioning event F and check P(F) > 0, (ii) compute P(E ∩ F), (iii) divide.
• Process — independence check: compute P(E), P(F), P(E ∩ F) separately; check whether the product equals the intersection probability. If yes, independent; if not, dependent.
• Process — total probability: identify a partition {Eᵢ}; tabulate P(Eᵢ) and P(A|Eᵢ); sum the products.
• Process — Bayes: identify the partition and the evidence A; compute total probability denominator; divide P(Eᵢ)·P(A|Eᵢ) by it.
• Process — tree diagram for sequential drawing: at each node, label branches by conditional probability; multiply along a path to get joint probability; add across paths to a leaf event.

2.4 Common confusions / NTA trap points

• Confusing independent with mutually exclusive — two events with nonzero probability cannot be both; NTA loves a distractor that says "if A and B are mutually exclusive they are independent" (NCERT §13.4, p. 418, Remark ii).
• Forgetting that P(A|B) requires P(B) ≠ 0; when P(B) = 0, P(A|B) is not defined (NCERT Exercise 13.1, Q16, p. 414 — option C is the correct one).
• Treating P(E|F) and P(F|E) as interchangeable — they are equal only when P(E) = P(F) (NCERT Exercise 13.1, Q17, p. 415).
• For three events, pairwise independence does not imply mutual independence — the triple product P(A ∩ B ∩ C) = P(A)·P(B)·P(C) must also hold (NCERT §13.4, p. 418, Remark iv).
• In Bayes' problems, mis-identifying the partition: the Eᵢ must be pairwise disjoint AND exhaustive AND have nonzero probability (NCERT §13.5.1, p. 423).
• Writing the total probability formula with the wrong conditioning direction: it is Σ P(Eⱼ)·P(A|Eⱼ), not Σ P(Eⱼ)·P(Eⱼ|A) (NCERT §13.5.2, p. 424).
• Forgetting that "without replacement" changes P(F|E); students sometimes carry over the with-replacement probability and over-count favourable outcomes.
• Mis-stating Property 2 by dropping the −P((A ∩ B)|F) term; valid only when A, B are disjoint within F.
• Confusing prior and posterior: P(Eᵢ) is prior (before seeing A), P(Eᵢ|A) is posterior (after seeing A). They are equal only when A is independent of Eᵢ.
• Mis-applying Bayes to non-partitions; if the Eᵢ overlap or are not exhaustive, the denominator does not equal P(A).
• Using addition rule P(A ∪ B) = P(A) + P(B) without subtracting P(A ∩ B), forgetting the inclusion-exclusion correction (a Class XI carryover error).
• Forgetting that mutually exclusive events with P(A), P(B) > 0 satisfy P(A ∩ B) = 0 ≠ P(A)·P(B), hence cannot be independent.
• Forgetting that conditional probability is a probability (lies in [0, 1]); negative or > 1 answers signal arithmetic error.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Exercise 13.1 Q1, p. 413). P(E) = 0.6, P(F) = 0.3, P(E ∩ F) = 0.2. Find P(E|F) and P(F|E).

Step 1 — apply definition: P(E|F) = P(E ∩ F)/P(F) = 0.2/0.3 = 2/3. Step 2 — apply other direction: P(F|E) = 0.2/0.6 = 1/3. Step 3 — observe asymmetry: P(E|F) ≠ P(F|E). Answer: 2/3 and 1/3.

Example B (NCERT Example 9, p. 416). Two cards drawn successively without replacement from a 52-card pack. Probability both are kings?

Step 1 — P(first king): 4/52. Step 2 — P(second king | first king): 3/51. Step 3 — multiply: (4/52)(3/51) = 12/2652 = 1/221. Answer: 1/221.

Example C (NCERT Example 10, p. 419). Die thrown; E = "multiple of 3", F = "even". Are E, F independent?

Step 1 — list sets: E = {3, 6}, F = {2, 4, 6}, E ∩ F = {6}. Step 2 — compute probabilities: P(E) = 2/6 = 1/3; P(F) = 3/6 = 1/2; P(E ∩ F) = 1/6. Step 3 — check product: P(E)·P(F) = (1/3)(1/2) = 1/6 = P(E ∩ F). Answer: Independent.

Example D (NCERT Example 15, p. 424–425). P(strike) = 0.65; P(complete | strike) = 0.32; P(complete | no strike) = 0.80. P(complete)?

Step 1 — partition: E₁ = strike, E₂ = no strike; P(E₁) = 0.65, P(E₂) = 0.35. Step 2 — apply total probability: P(A) = 0.65(0.32) + 0.35(0.80) = 0.208 + 0.280 = 0.488. Step 3 — state: P(complete on time) = 0.488. Answer: 0.488.

Example E (NCERT Example 19, pp. 428–429). Machines A, B, C produce 25%, 35%, 40% of bolts; defective rates 5%, 4%, 2%. Given a defective, P(from B)?

Step 1 — priors and likelihoods: P(A) = 0.25, P(B) = 0.35, P(C) = 0.40; P(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02. Step 2 — total probability of defective: P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345. Step 3 — apply Bayes: P(B|D) = (0.35 · 0.04)/0.0345 = 0.0140/0.0345 = 28/69. Answer: 28/69.

Example F (NCERT Example 17, p. 427). A box has 3 coins: one two-headed, one biased (P(H) = 0.75), one fair. A coin is picked at random and tossed; it shows head. What is the probability it is the two-headed coin?

Step 1 — partition: E₁ = two-headed, E₂ = biased, E₃ = fair; each prior = 1/3. Likelihoods P(H|E₁) = 1, P(H|E₂) = 3/4, P(H|E₃) = 1/2. Step 2 — total probability of head: P(H) = (1/3)(1 + 3/4 + 1/2) = (1/3)(9/4) = 3/4. Step 3 — Bayes: P(E₁|H) = (1/3 · 1)/(3/4) = 4/9. Answer: 4/9.