Probability

Probability

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Probability quantifies how likely an event is, expressed as a number from 0 to 1. The classical formula P(E) = n(E)/n(S) counts favourable outcomes over the total equally likely outcomes in the sample space S. Three identities dominate the CUET UG paper: the complement rule, the addition rule, and Bayes’ theorem.

• Complement: P(E) + P(E’) = 1
• Addition: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); if mutually exclusive, P(A ∩ B) = 0
• Multiplication/Conditional: P(A ∩ B) = P(A)·P(B|A); for independent events P(A ∩ B) = P(A)·P(B)
• Bayes’: P(Aᵢ|B) = P(B|Aᵢ)·P(Aᵢ) / Σ P(B|Aₖ)·P(Aₖ)
• Odds trap: “odds in favour” of E = n(E) : n(E’), not P(E)
• Independent vs mutually exclusive: only one can be true when both P(A), P(B) > 0

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Classical and Axiomatic Probability

Classical probability assumes every elementary outcome in S is equally likely, giving P(E) = n(E)/n(S). The axiomatic version (Kolmogorov) requires 0 ≤ P(E) ≤ 1, P(S) = 1, and additivity over disjoint events. CUET problems typically use classical probability backed by combinations — for example, drawing 2 cards from 52 uses C(52,2) as n(S).

Addition and the Union Rule

For any two events A and B, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive (no common outcome), the intersection term vanishes. A common three-event error is double-subtracting: the correct form is P(A∪B∪C) = ΣP(Aᵢ) − ΣP(Aᵢ∩Aⱼ) + P(A∩B∩C).

Conditional Probability and Independence

P(A|B) = P(A ∩ B)/P(B), valid only when P(B) ≠ 0. Independence means the occurrence of B does not change A’s likelihood, so P(A|B) = P(A) and equivalently P(A ∩ B) = P(A)·P(B). Mutually exclusive events with positive probabilities are never independent — this distinction is a CUET favourite.

Worked Pattern: Cards

Two cards drawn without replacement: P(both kings) = (4/52)·(3/51) = 1/221. With replacement: (4/52)² = 1/169.

Concept	Formula / Key fact
Classical probability	P(E) = n(E)/n(S), equally likely outcomes only
Complement rule	P(E’) = 1 − P(E)
Addition rule	P(A∪B) = P(A) + P(B) − P(A∩B)
Multiplication rule	P(A∩B) = P(A)·P(B|A)
Independence check	P(A∩B) = P(A)·P(B)
Total probability	P(B) = Σ P(B|Aᵢ)·P(Aᵢ) over partition {Aᵢ}

• Bayes’ theorem reverses a conditional: given P(B|Aᵢ) and prior P(Aᵢ), find P(Aᵢ|B).
• Tree diagrams partition the sample space and turn Bayes’ problems into arithmetic.
• “Odds in favour of E” = n(E) : n(E’); converting requires P(E) = odds/(1+odds).

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Law of Total Probability and Bayes’ Theorem

The law partitions S into mutually exclusive, exhaustive events {A₁, A₂, …, Aₙ} with ΣP(Aᵢ) = 1, giving P(B) = Σ P(B|Aᵢ)·P(Aᵢ). Bayes’ theorem then computes the posterior P(Aᵢ|B) = P(B|Aᵢ)·P(Aᵢ)/P(B). This two-step structure — total probability as denominator, likelihood × prior in numerator — appears in roughly one of the 2–4 CUET Probability questions.

Edge Cases and Common Traps

Biased coins and loaded dice break the classical assumption; only the axiomatic definition holds. “At least one” problems are usually fastest via complement: P(at least one six in 3 dice) = 1 − (5/6)³. Conditional probability questions frequently hide the order of drawing; “without replacement” changes the denominator between successive draws.

Common Mistakes

• Applying P(E) = n(E)/n(S) when outcomes are not equally likely.
• Treating “odds in favour” as P(E).
• Dropping the denominator in Bayes’ theorem.
• Confusing P(A∪B) for independent events: here P(A∩B) = P(A)P(B) ≠ 0 in general.

Worked Micro-Example

A factory has three machines producing 50%, 30%, 20% of output with defect rates 2%, 3%, 5%. P(defective) = 0.5·0.02 + 0.3·0.03 + 0.2·0.05 = 0.031. P(M₁|defect) = (0.5·0.02)/0.031 ≈ 0.323.

Mistake	Correction
Adding P(A)+P(B) always	Subtract P(A∩B) unless events are mutually exclusive
Odds = probability	Odds in favour of E = n(E)/n(E’), then P = odds/(1+odds)
Classical formula on biased setup	Use axiomatic definition; outcomes not equally likely
Bayes’ without denominator	Denominator is P(B) via total probability law

1. A bag has 5 red, 3 blue balls. Two are drawn without replacement. Find P(one red, one blue).
2. Given P(A)=0.4, P(B|A)=0.5, P(B|A’)=0.2, compute P(A|B).

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