Probability

🟢 Lite — Quick Review

Rapid summary for last-minute revision before your exam.

Probability — Key Facts for JEE Main Sample space (S): set of all possible outcomes Event: subset of sample space P(E) = n(E)/n(S) for equally likely outcomes; always 0 ≤ P(E) ≤ 1 Conditional probability: P(A|B) = P(A ∩ B)/P(B) (probability of A given B has occurred) Independent events: P(A ∩ B) = P(A)·P(B); if P(A|B) = P(A) Mutually exclusive: P(A ∪ B) = P(A) + P(B); P(A ∩ B) = 0 ⚡ Exam tip: Bayes’ theorem is frequently tested in JEE Main — P(A|B) = P(B|A)·P(A) / P(B)!

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🟡 Standard — Core Study

Standard content for students with a few days to months.

Probability — JEE Main Study Guide

Complement rule: P(A’) = 1 − P(A); useful for “at least one” problems Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) For mutually exclusive: P(A ∪ B) = P(A) + P(B) Bayes’ theorem: If E₁, E₂, …, E_n are partition of S, then P(E_i|A) = P(E_i)·P(A|E_i) / Σ P(E_j)·P(A|E_j) Total probability: P(A) = Σ P(E_i)·P(A|E_i) for partition {E_i}

Counting approach to probability:

• When outcomes are equally likely: P(event) = favourable outcomes / total outcomes
• For complex counting problems: use P&C techniques to find numerator and denominator

Key distributions:

• Binomial distribution: P(X = r) = nC_r · p^r · q^{n−r} where p = success prob, q = 1−p
• Mean of binomial = np; variance = npq

Conditional probability with cards/dice: P(drawing second ace given first was ace) = 3/51 = 1/17 (without replacement) P(second ace given first was ace) = 4/52 × 3/51 (conditional, multiply)

Independent vs mutually exclusive:

• Two events A and B are independent if P(A ∩ B) = P(A)·P(B)
• Mutually exclusive: cannot both occur; P(A ∩ B) = 0
• Independent and mutually exclusive can both be true only if one has probability 0

Common mistakes:

• Confusing “with replacement” vs “without replacement” — probabilities change

• Treating non-independent events as independent

• Forgetting to subtract P(A ∩ B) in union formula

• Key formula: P(A|B) = P(A ∩ B)/P(B); P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

• Common trap: P(A|B) ≠ P(B|A) — be careful which is in numerator! Bayes’ theorem swaps these

• Exam weight: 1–2 questions per year (4–8 marks); frequently combined with P&C for counting probability

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🔴 Extended — Deep Dive

Comprehensive coverage for students on a longer study timeline.

Probability — Comprehensive JEE Main Notes

Bayes’ theorem derivation: Let E₁, E₂ partition S. For any event A: P(E_i|A) = P(E_i ∩ A)/P(A) = P(E_i)·P(A|E_i) / Σ P(E_j)·P(A|E_j)

Application: diagnostic test problem type: P(Disease|+test) = P(+test|Disease)·P(Disease) / [P(+test|Disease)·P(Disease) + P(+test|No Disease)·P(No Disease)] This is the Bayesian calculation — very common in JEE Main

Maximum likelihood estimation: The value that maximises P(X=x) for binomial is the integer part of (n+1)p

Geometric probability: When outcomes form a continuous range: P(event) = length of favourable region / total length Example: Two numbers chosen from [0,1], probability their sum < 1 = area of triangle in unit square = 1/2

Derangements (probability perspective): Probability that no one gets their own hat in hat-check problem: D_n/n! = 1 − 1/1! + 1/2! − 1/3! + … + (−1)^n/n! As n → ∞, this approaches 1/e ≈ 0.3679

Important identities:

• P(A’ ∩ B) = P(B) − P(A ∩ B)
• P(A ∩ B’) = P(A) − P(A ∩ B)
• P((A ∪ B)’) = 1 − P(A ∪ B) = P(A’ ∩ B’)
• P(A ∩ B) = P(A) + P(B) − P(A ∪ B)

Random variable concepts: For discrete random variable X with values x_i and probabilities p_i:

• E(X) = Σ p_i x_i (expectation/mean)
• Var(X) = E(X²) − [E(X)]²
• For binomial: E(X) = np, Var(X) = npq

Odds:

• Odds in favour of event E: P(E)/P(E’) = a:b
• Probability from odds: P(E) = a/(a+b)

Conditional probability with urn problems: Drawing from urn without replacement: each draw changes remaining composition P(first red and second blue) = (r/(r+b)) × (b/(r+b−1))

Solved problem type: “From a pack of 52 cards, 3 are drawn without replacement. Find probability all are kings.” = 4C₃/52C₃ = 4/22100 = 1/5525 Or using sequential: (4/52)×(3/51)×(2/50) = 24/132600 = 1/5525 ✓

Solved problem type: “Probability that at least one die shows 6 when rolling n dice = 1 − (5/6)^n” This uses complement: none show 6 = (5/6)^n

Probability of getting exactly r heads in n tosses: P(r heads) = nC_r · (1/2)^n (fair coin) For biased coin with P(H) = p: nC_r · p^r · (1−p)^{n−r}

JEE Main special patterns:

• “Two cards drawn from pack. Probability both are spades?” = 13C₂/52C₂ = 3/51 = 1/17

• “At least one head in 3 tosses?” = 1 − P(no heads) = 1 − (1/2)³ = 7/8

• “Draw 3 balls from bag with 4 white, 5 black. Probability exactly 2 white?” = C(4,2)·C(5,1)/C(9,3)

• Remember: P(A|B) = P(A∩B)/P(B); for independent events, P(A∩B) = P(A)·P(B); complement for “at least one”: 1 − P(none); for “without replacement”, multiply sequentially using updated counts

• Previous years: “A bag contains 5 white and 7 black balls. Probability of drawing 2 white and 1 black without replacement” [2023]; “Bayes’ theorem: box1 has 2 white 4 black, box2 has 3 white 3 black. Find P(box1|first ball is white)” [2024]

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📊 JEE Main Exam Essentials

Detail	Value
Questions	90 (30 per subject)
Time	3 hours
Marks	300 (90 per subject)
Section	Physics (30), Chemistry (30), Mathematics (30)
Negative	−1 for wrong answer
Mode	Computer-based

🎯 High-Yield Topics for JEE Main Mathematics

• Calculus (Differentiation + Integration) — ~35 marks combined
• Coordinate Geometry (straight lines, circles, conics) — ~20 marks
• Algebra (Complex Numbers, Quadratics, P&C, Probability) — ~25 marks
• Trigonometry + Inverse Trigonometry — ~15 marks
• Vector + 3D — ~15 marks

📝 Previous Year Question Patterns

• Probability: 1–2 questions per year, 4–8 marks
• Common patterns: conditional probability, Bayes’ theorem, binomial distribution, urn problems
• Weight: medium-high frequency, high difficulty but scoring

💡 Pro Tips

• Bayes’ theorem is a JEE Main favourite — understand the partition concept thoroughly
• For “at least one” problems, complement is always easier: P(at least one) = 1 − P(none)
• When drawing without replacement, multiply sequentially with updated counts
• For geometric probability, draw the region and compute area/length ratios
• Always check whether events are independent or mutually exclusive — different formulas
• When the problem says “given that”, it’s a conditional probability problem
• Odds can be converted: if odds in favour are a:b, then P = a/(a+b)

🔗 Official Resources

• NTA Official JEE Main
• JEE Main Syllabus PDF

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