🎯 Key Points
- A random experiment has more than one possible outcome and none can be predicted with certainty; the set of all outcomes is the sample space S
- An event is any subset of S. The empty set ∅ is the impossible event and S itself is the sure event
- Axioms: P(A) ≥ 0, P(S) = 1, and for mutually exclusive events P(A ∪ B) = P(A) + P(B)
- For equally likely outcomes, P(A) = n(A)/n(S) — favourable outcomes over total outcomes
- Addition rule in general: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); the last term vanishes only when A and B are mutually exclusive
- Complement: P(A′) = 1 − P(A) — often far quicker than counting the event directly
- Probability always lies in [0, 1]; a negative value or one above 1 signals an arithmetic slip
- Mutually exclusive (A ∩ B = ∅) is not the same as exhaustive (A ∪ B = S)
Random Experiments and Sample Space
An experiment is called random if it has more than one possible outcome and the outcome cannot be predicted in advance. Every possible outcome is a sample point, and the set of all of them is the sample space, written S.
- Tossing one coin → S = {H, T}, so n(S) = 2
- Tossing two coins → S = {HH, HT, TH, TT}, so n(S) = 4
- Rolling one die → S = {1, 2, 3, 4, 5, 6}, so n(S) = 6
- Rolling two dice → n(S) = 36
- Drawing one card from a standard pack → n(S) = 52
Types of Events
- Impossible event — ∅, probability 0
- Sure (certain) event — S, probability 1
- Simple event — a single sample point
- Compound event — more than one sample point
- Complementary event A′ — everything in S that is not in A
- Mutually exclusive — A ∩ B = ∅; the two cannot happen together
- Exhaustive — the events together cover all of S
Axiomatic Probability
A probability function P assigns a number to each event subject to three axioms:
- P(A) ≥ 0 for every event A
- P(S) = 1
- If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B)
Everything else follows. In particular P(∅) = 0, P(A′) = 1 − P(A), and 0 ≤ P(A) ≤ 1.
The Addition Rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Subtracting P(A ∩ B) prevents double-counting the overlap. If the events are mutually exclusive the overlap is empty and the rule collapses to P(A) + P(B). For three events:
P(A∪B∪C) = P(A)+P(B)+P(C) − P(A∩B) − P(B∩C) − P(A∩C) + P(A∩B∩C)
💡 Advanced Edge
- Phrases like "at least one" are usually fastest through the complement: P(at least one) = 1 − P(none).
- "Equally likely" is an assumption, not a guarantee. The formula n(A)/n(S) is only valid when every sample point genuinely has the same chance — a loaded die breaks it.
- Odds differ from probability: odds in favour of A are P(A) : P(A′). Odds of 3 : 2 mean a probability of 3/5, not 3/2.
- When two dice are rolled, outcomes like (2, 3) and (3, 2) are distinct sample points. Treating the pair as unordered gives 21 outcomes that are not equally likely — a classic trap.
Example 1 — Two dice
Q: Two dice are thrown. Find the probability that the sum is 8.
Step 1 — Total outcomes: n(S) = 6 × 6 = 36.
Step 2 — Favourable: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes.
Answer: P = 5/36.
Example 2 — Using the addition rule
Q: A card is drawn from a well-shuffled pack. Find the probability that it is a king or a heart.
Step 1 — P(king) = 4/52, P(heart) = 13/52.
Step 2 — Overlap: the king of hearts, so P(king ∩ heart) = 1/52.
Step 3 — Apply the rule: 4/52 + 13/52 − 1/52 = 16/52.
Answer: 4/13.
Example 3 — "At least one" via the complement
Q: Three coins are tossed. Find the probability of getting at least one head.
Step 1 — n(S) = 2³ = 8.
Step 2 — The complement of "at least one head" is "no heads", i.e. TTT — a single outcome.
Step 3 — P(no head) = 1/8.
Answer: P(at least one head) = 1 − 1/8 = 7/8. Note: counting the 7 favourable cases directly works too, but the complement is faster and less error-prone.