📚 StudyHub

📐 Mathematics  ·  Class 12  ·  JEE

Probability

Chance, events, conditional probability, and Bayes theorem

Practice Probability Quiz — 100% Free →
Reading time~7 min
Revision time~2 min
Last updated2026-07-17
1 Read the chapter ~7 min

🎯 Key Points

  • P(A∪B)=P(A)+P(B)-P(A∩B) always; for MUTUALLY EXCLUSIVE events only, P(A∩B)=0 so it simplifies to P(A)+P(B)
  • Independent events: P(A∩B)=P(A)×P(B) — this is a SPECIAL case, not the general multiplication law, which is P(A∩B)=P(A)×P(B|A)
  • Bayes' theorem reverses a conditional: given P(A|B), find P(B|A) — essential whenever a problem gives "probability of evidence given hypothesis" but asks for "probability of hypothesis given evidence"
  • Binomial: mean=np, variance=npq (always less than mean since q<1); Poisson: mean=variance=λ — the EQUAL mean/variance is the signature that distinguishes Poisson from binomial

Probability

Probability measures the likelihood of events occurring, ranging from 0 (impossible) to 1 (certain).

Basic Definitions

  • Sample space (S): Set of all possible outcomes
  • Event (E): Subset of sample space
  • P(E) = n(E)/n(S) (classical definition)
  • 0 ≤ P(E) ≤ 1; P(S) = 1; P(∅) = 0

Addition Law

UABA and BA onlyB onlyOutside both circles: complement of (A union B)

Venn diagram of the universal set U with events A and B, showing the intersection (A and B), the parts unique to each event, and the complement region outside both.

  • P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
  • Mutually exclusive events: P(A ∪ B) = P(A) + P(B) [since P(A ∩ B) = 0]
  • Complement: P(A') = 1 - P(A)

Conditional Probability

  • P(A|B) = P(A ∩ B) / P(B)
  • Independent events: P(A ∩ B) = P(A) × P(B); P(A|B) = P(A)
  • Multiplication law: P(A ∩ B) = P(A) × P(B|A)

Bayes' Theorem

P(B|A) = P(A|B) × P(B) / P(A): used to update probability given new evidence.

Probability Distributions

  • Binomial: P(X=r) = nCr × pʳ × qⁿ⁻ʳ (n trials, p success, q=1-p). Mean = np, Variance = npq
  • Poisson: P(X=r) = e^(-λ) × λʳ/r! (rare events). Mean = Variance = λ
  • Normal: Continuous, bell-shaped. Standardize: Z = (X-μ)/σ

Common Experiments

  • Dice (6 outcomes): P(even) = 3/6 = 1/2
  • Coin (2 outcomes): P(head) = 1/2
  • Cards (52): 4 suits × 13 ranks; P(ace) = 4/52 = 1/13

🚀 JEE Advanced Edge

Total probability theorem as the engine behind Bayes' theorem: If B₁,B₂,...,Bₙ partition the sample space (mutually exclusive, exhaustive), then P(A) = ΣP(Bᵢ)·P(A|Bᵢ) — this "law of total probability" is what computes the denominator P(A) inside Bayes' theorem when it isn't given directly, making the two theorems a matched pair rather than independent tools.

Why "at least one" probability problems are solved via the complement: Computing P(at least one success in n trials) directly requires summing P(exactly 1)+P(exactly 2)+...+P(exactly n) — tedious. Instead, P(at least one) = 1 - P(none), and P(none) is a single easy term to compute, making the complement approach dramatically faster whenever a problem contains the phrase "at least one."

Worked problem: Three machines A, B, C produce 25%, 35%, 40% of a factory's output respectively, with defect rates 5%, 4%, 2%. A randomly selected item is found defective. Find the probability it came from machine A. Approach: By total probability, P(defective) = 0.25×0.05 + 0.35×0.04 + 0.40×0.02 = 0.0125+0.014+0.008 = 0.0345. By Bayes: P(A|defective) = P(A)×P(defective|A)/P(defective) = 0.0125/0.0345 ≈ 0.362 (about 36.2%).

Worked Example: Conditional Probability

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.

P(1st red) = 5/8. Given 1st is red, only 4 red remain out of 7 balls: P(2nd red | 1st red) = 4/7.

P(both red) = (5/8) × (4/7) = 20/56 = 5/14. Without replacement: the denominator decreases by 1 for each draw, and the number of favourable outcomes also changes.

Worked Example: Addition Rule (Mutually Non-exclusive Events)

A card is drawn from a deck of 52. Find P(red or face card).

P(red) = 26/52, P(face card) = 12/52, P(red AND face card) = 6/52 (6 red face cards). By addition rule: P(red ∪ face) = 26/52 + 12/52 − 6/52 = 32/52 = 8/13. Subtract the intersection to avoid double-counting.

Multiplication Theorem and Independent Events

  • Multiplication theorem: P(A ∩ B) = P(A)·P(B|A) = P(B)·P(A|B), valid whenever P(A) and P(B) are non-zero.
  • For three events: P(A ∩ B ∩ C) = P(A)·P(B|A)·P(C | A ∩ B).
  • Independent events: A and B are independent if P(A ∩ B) = P(A)·P(B); equivalently P(A|B) = P(A) — one event's occurrence does not change the other's probability.
  • Independent is NOT the same as mutually exclusive: two events with non-zero probability cannot be both (mutually exclusive means P(A ∩ B) = 0).
  • If A and B are independent, so are A' and B, A and B', and A' and B'.

Total Probability Theorem

If E₁, E₂, …, Eₙ form a partition of the sample space (mutually exclusive, exhaustive, each with non-zero probability) and A is any event, then:

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + … + P(Eₙ)·P(A|Eₙ)

This computes the overall probability of A by conditioning on which case Ei occurred, and supplies the denominator P(A) used inside Bayes' theorem.

Random Variable, Mean and Variance

  • A random variable X is a real-valued function on the sample space; a probability distribution lists each value xᵢ with its probability pᵢ, where every pᵢ ≥ 0 and Σpᵢ = 1.
  • Mean (expectation): μ = E(X) = Σ xᵢ pᵢ.
  • Variance: Var(X) = Σ (xᵢ − μ)² pᵢ = E(X²) − [E(X)]², where E(X²) = Σ xᵢ² pᵢ.
  • Standard deviation: σ = √Var(X) (measures spread about the mean).

Bernoulli Trials and Binomial Distribution

  • Bernoulli trials: independent, repeated trials with exactly two outcomes (success/failure) and the same success probability p on every trial.
  • Binomial distribution: for n trials, P(X = r) = nCr pr qn−r, with q = 1 − p and r = 0, 1, …, n.
  • Mean = np, Variance = npq; the variance is always less than the mean because q < 1.
  • Written as B(n, p); the terms are the successive terms of the expansion of (q + p)n.
2 Revise ~2 min before the exam

📐 Formula Sheet

  • Classical definition: P(E) = favourable outcomes / total outcomes, 0 ≤ P(E) ≤ 1
  • Complement: P(E') = 1 − P(E)
  • Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); for mutually exclusive events the last term is 0
  • Conditional: P(A|B) = P(A ∩ B)/P(B)
  • Multiplication: P(A ∩ B) = P(A)·P(B|A); independent events ⇒ P(A ∩ B) = P(A)P(B)
  • Total probability: P(A) = ΣP(Ei)·P(A|Ei)
  • Bayes' theorem: P(Ei|A) = P(Ei)P(A|Ei) / ΣP(Ej)P(A|Ej)
  • Binomial distribution: P(X = r) = ⁿCrprqn−r, mean = np, variance = npq
  • Expectation: E(X) = Σxipi  |  Var(X) = E(X²) − [E(X)]²
3 Practice apply it

✍️ Worked Examples

Example 1 — Conditional probability
Q: Two dice are thrown. Given that the sum is 7, what is the probability that one die shows a 2?
Step 1 — List outcomes with sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — six in all.
Step 2 — Of these, which contain a 2? (2,5) and (5,2) — two outcomes.
Step 3 — Conditional probability: 2/6 = 1/3.
Answer: 1/3. Key idea: conditioning shrinks the sample space from 36 to just those 6 outcomes.

Example 2 — Bayes' theorem
Q: Factory A makes 60% of items with a 2% defect rate; factory B makes 40% with a 5% defect rate. An item is found defective. What is the probability it came from A?
Step 1 — Write the knowns: P(A) = 0.6, P(D|A) = 0.02, P(B) = 0.4, P(D|B) = 0.05.
Step 2 — Total probability of a defect: P(D) = (0.6)(0.02) + (0.4)(0.05) = 0.012 + 0.020 = 0.032.
Step 3 — Bayes: P(A|D) = (0.6)(0.02)/0.032 = 0.012/0.032.
Answer: 0.375, i.e. 37.5%. Note: although A makes most items, its lower defect rate means most defective ones come from B.

Example 3 — Binomial distribution
Q: A fair coin is tossed 5 times. Find the probability of exactly 3 heads.
Step 1 — Identify: n = 5, r = 3, p = q = ½.
Step 2 — Apply the formula: P = ⁵C₃(½)³(½)² = 10 × (1/8) × (1/4).
Step 3 — Compute: P = 10/32 = 5/16.
Answer: 5/16 = 0.3125. Note: mean = np = 2.5, so 3 heads is close to the most likely outcome.

Practice Probability Quiz — 100% Free →

Frequently Asked Questions — Probability

What are the key concepts in Probability?
Chance, events, conditional probability, and Bayes theorem
Is Probability important for JEE?
Yes. Probability is part of the Mathematics Class 12 NCERT syllabus and is directly tested in JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Probability questions on StudyHub?
Open StudyHub and select Mathematics → Probability. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at JEE level with full step-by-step explanations.

References

  1. NCERT Class 12 Mathematics Textbook — Chapter: Probability
  2. CBSE Curriculum — Mathematics (Class 12)
  3. NTA JEE Main Official Syllabus — subject-wise topic list