6 Statistics: Data, Sampling & Presentation · 6.1 Data, sampling, probability and presentation
6.1.3 Probability: classical, relative-frequency and axiomatic definitions; addition and multiplication theorems
Probability measures the chance of an event on a scale from 0 (impossible) to 1 (certain); it is defined classically (favourable ÷ equally likely cases), by long-run relative frequency, or by Kolmogorov's axioms, and is combined with the addition (or) and multiplication (and) theorems.
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Core facts
- Random experiment: outcome uncertain, all possible outcomes known (tossing a coin).
- Sample space (S): set of all possible outcomes. Event: a subset of S.
- Equally likely: no outcome is preferred. Mutually exclusive: cannot happen together (A ∩ B = ∅). Exhaustive: together they cover all outcomes.
- Independent events: occurrence of one does not change the chance of the other: P(A ∩ B) = P(A) P(B).
- Classical (a priori) definition (Laplace): P(A) = m/n = favourable cases ÷ total equally likely, mutually exclusive, exhaustive cases.
- Relative frequency (empirical) definition (von Mises): P(A) = limit of m/n as the number of trials n → ∞.
- Axiomatic definition (A.N. Kolmogorov, 1933):
- P(A) ≥ 0 for every event A;
- P(S) = 1;
- for mutually exclusive events, P(A ∪ B ∪ …) = P(A) + P(B) + ….
- Range: 0 ≤ P(A) ≤ 1. Complement: P(A′) = 1 − P(A). P(∅) = 0.
- Odds in favour of A = a : b means P(A) = a/(a + b); odds against = b : a.
- Addition theorem:
- mutually exclusive: P(A ∪ B) = P(A) + P(B);
- any two events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- Multiplication theorem:
- independent: P(A ∩ B) = P(A) × P(B);
- general: P(A ∩ B) = P(A) × P(B | A).
- Conditional probability: P(B | A) = P(A ∩ B) ÷ P(A), with P(A) > 0.
Explained
1. Why three definitions.
| Definition | Idea | Limitation |
|---|---|---|
| Classical | count equally likely cases | fails when cases are not equally likely or are infinite |
| Relative frequency | long-run proportion | needs repeated trials; the limit may not be observable |
| Axiomatic | rules any probability must obey | does not say how to find the value |
2. Standard counts. Coin: 2 outcomes; die: 6; two dice: 36; three coins: 8; pack of cards: 52 (4 suits of 13; 12 face cards; 4 aces; 26 red).
Example 1. Two dice are thrown. Find the probability that the sum is 7. Solution. Favourable pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 of 36. Answer: 1/6
Example 2. A card is drawn from a pack of 52. Find P(king or heart). Solution. Not mutually exclusive (king of hearts). 4/52 + 13/52 − 1/52 = 16/52. Answer: 4/13
Example 3. Three coins are tossed. Find P(at least one head). Solution. P(no head) = 1/8, so 1 − 1/8. Answer: 7/8
Example 4. Odds in favour of an event are 3 : 5. Find its probability. Solution. 3/(3 + 5). Answer: 3/8
3. Independent events.
Example 5. P(A) = 0.5, P(B) = 0.4, A and B independent. Find P(A ∪ B). Solution. P(A ∩ B) = 0.2. 0.5 + 0.4 − 0.2. Answer: 0.7 (Check: 1 − P(A′)P(B′) = 1 − 0.5 × 0.6 = 0.7.)
Example 6. A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4. Find the probability that it is solved. Solution. P(none solves) = (1/2)(2/3)(3/4) = 1/4. Solved = 1 − 1/4. Answer: 3/4
4. Dependent events (without replacement).
Example 7. Two cards are drawn one after the other without replacement. Find P(both aces). Solution. (4/52) × (3/51) = 12/2652. Answer: 1/221
Deeper
Mutually exclusive is not independent. If P(A) > 0 and P(B) > 0 and A, B are mutually exclusive, then P(A ∩ B) = 0 ≠ P(A)P(B): they are dependent (one happening rules out the other).
Three events (addition): P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C).
If A and B are independent, so are A and B′, A′ and B, and A′ and B′.
Bayes' theorem. If E₁, …, Eₙ are mutually exclusive and exhaustive and A is any event, P(Eᵢ | A) = P(Eᵢ) P(A | Eᵢ) ÷ Σ P(Eⱼ) P(A | Eⱼ). It revises prior probabilities into posterior probabilities after evidence A.
Example 8. Urn I has 3 white and 2 black balls; urn II has 2 white and 3 black. An urn is chosen at random and a ball drawn is white. Find P(it came from urn I). Solution. Numerator 0.5 × 3/5 = 0.3. Denominator 0.3 + 0.5 × 2/5 = 0.5. 0.3/0.5. Answer: 3/5 (0.6)
Example 9. Find the probability that a leap year chosen at random has 53 Sundays. Solution. 366 days = 52 weeks + 2 days. The 2 extra days are one of 7 equally likely pairs (Sun–Mon, Mon–Tue, …, Sat–Sun); 2 contain Sunday. Answer: 2/7
Example 10. Which statement is true? (a) If P(A) = 0.6 and P(B) = 0.7, A and B can be mutually exclusive. (b) The axiomatic approach was given by Laplace. (c) P(A ∪ B) ≤ P(A) + P(B) always. (d) Independent events with positive probabilities are mutually exclusive. Solution. (a) 0.6 + 0.7 > 1, impossible. (b) Kolmogorov. (d) false. (c) follows from the addition theorem since P(A ∩ B) ≥ 0. Answer: (c)
Common traps
- ✗ "Mutually exclusive events are independent." ✓ With positive probabilities they are dependent; P(A ∩ B) = 0 ≠ P(A)P(B).
- ✗ "Odds in favour 3 : 5 means P = 3/5." ✓ P = 3/(3 + 5) = 3/8.
- ✗ "P(A or B) = P(A) + P(B) always." ✓ Subtract P(A ∩ B) unless they are mutually exclusive.
- ✗ "The axiomatic definition is due to von Mises." ✓ Kolmogorov (1933); von Mises gave the relative frequency approach; Laplace the classical.
- ✗ "Two dice give 12 outcomes." ✓ 36 ordered outcomes.
- ✗ "Probability can be 1.2 for a very likely event." ✓ It lies between 0 and 1.
- ✗ "Without replacement, the second draw has the same probability." ✓ Use P(B | A): the pack has changed.
Quick revision
- 0 ≤ P(A) ≤ 1; P(A′) = 1 − P(A).
- Classical (Laplace): m/n with equally likely cases.
- Relative frequency (von Mises): limit of m/n.
- Axiomatic (Kolmogorov 1933): non-negativity, P(S) = 1, additivity.
- Odds a : b in favour → P = a/(a + b).
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- Independent: P(A ∩ B) = P(A)P(B).
- General: P(A ∩ B) = P(A) P(B | A).
- At least one = 1 − P(none).
- Bayes: posterior ∝ prior × likelihood.
Practise this topic. Questions on Probability: definitions, with answers explained.
All Data, sampling, probability and presentation notes · APPSC Paper I syllabus · APPSC Paper I Mock 1 with answers · Start free
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