Big picture
A fair coin toss has an uncertain outcome, but its possible outcomes and their probabilities can be described precisely. Probability provides a mathematical model of uncertainty. It helps distinguish a possible event from a certain one and a large number of favourable outcomes from a genuinely high chance.
Classical probability begins by constructing an appropriate sample space and checking whether its elementary outcomes are equally likely. Counting is useful only after that modelling step. For example, the sums from two dice are not equally likely even though the ordered pairs of die faces are equally likely for independent fair dice.
Learning goals
Describe random experiments and sample spaces.
Represent events and operations on events using sets.
Apply probability axioms and complement rules.
Use addition laws for overlapping and disjoint events.
Count equally likely outcomes accurately in multistep experiments.
Core ideas
1. Random experiments and sample spaces
A random experiment has well-defined possible outcomes but an unpredictable individual result under the model. The sample space, S, is the set of all elementary outcomes. An event is a subset of S. For one die, S = {1, 2, 3, 4, 5, 6}; the event of an even result is {2, 4, 6}.
For two coins, record outcomes in a fixed order: S = {HH, HT, TH, TT}. HT and TH are different elementary outcomes because the first and second positions differ. For two dice, ordered pairs (i, j), with each coordinate from 1 to 6, give 36 outcomes. If only the sum is recorded, the sample space changes, but the 11 possible sums do not acquire equal probabilities merely because they are listed once each.
2. Event types
A simple event contains one elementary outcome; a compound event contains more than one. The sure event is S and the impossible event is the empty set ∅. The complement A′ contains all outcomes in S not belonging to A.
For a die, let A = {2, 4, 6}. Then A′ = {1, 3, 5}. The statement A or A′ must occur is certain, while A and A′ cannot occur together. “At least one head” in two tosses means {HH, HT, TH}; “exactly one head” means {HT, TH}. These phrases describe different sets and therefore different probabilities.
3. Operations on events
Union, A ∪ B, means A or B or both. Intersection, A ∩ B, means both A and B. Difference, A − B, means A occurs but B does not. Events are mutually exclusive if A ∩ B = ∅; they are exhaustive if their union covers S.
For a die, let A be even and B be greater than 3. Then A = {2, 4, 6}, B = {4, 5, 6}, A ∩ B = {4, 6} and A ∪ B = {2, 4, 5, 6}. Thus “even or greater than 3” includes four faces, not six. “Not A and not B” is the complement of the union, whereas “not both A and B” is the complement of the intersection. A Venn diagram makes this distinction visible.
4. Axiomatic probability
Probability assigns a number P(A) to each event with P(A) ≥ 0, P(S) = 1, and additivity for mutually exclusive events. These imply P(∅) = 0 and 0 ≤ P(A) ≤ 1. An event cannot have probability 1.2 or −0.1.
For equally likely finite outcomes, P(A) = n(A)/n(S). Fairness and the experimental design justify equal likelihood; the formula is not a definition that makes arbitrary listed categories equal. If a biased coin has P(H) = 0.7, its two outcomes remain a sample space but are not assigned probability 1/2 each. In every valid finite model, elementary probabilities sum to 1.
5. Complement rule
Since A and A′ are disjoint and exhaustive, P(A′) = 1 − P(A). This is especially useful for “at least one” questions. With three independent fair coin tosses, there are 2³ = 8 equally likely sequences. Only TTT has no heads, so P(at least one head) = 1 − 1/8 = 7/8.
For two fair dice, P(at least one six) is easier through the complement. There are 5 × 5 = 25 ordered pairs containing no six out of 36, so probability = 1 − 25/36 = 11/36. Directly adding 6 outcomes with a first-die six and 6 with a second-die six double-counts (6, 6); subtracting that overlap also gives 11 favourable outcomes.
6. Addition law
For any events A and B, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The overlap is subtracted because it was counted twice. Only when events are mutually exclusive does this reduce to P(A) + P(B).
In the die example, P(A) = 3/6, P(B) = 3/6 and P(A ∩ B) = 2/6, so P(A ∪ B) = 3/6 + 3/6 − 2/6 = 2/3. If P(A) = 0.55, P(B) = 0.40 and P(A ∩ B) = 0.20, then P(A ∪ B) = 0.75 and P(neither) = 0.25. “Exactly one of A and B” excludes the intersection from both counts, giving P(A) + P(B) − 2P(A ∩ B) = 0.55.
Mutually exclusive does not mean independent. Two distinct faces on one die cannot occur together, whereas independent events concern how one occurrence affects another's probability. Formal conditional probability is studied later; do not import an unjustified independence assumption to simplify a union.
7. Counting cards and selections
A standard 52-card deck has four suits of 13 cards, with 12 face cards: jack, queen and king in each suit. Aces are not included as face cards in this convention. A single randomly selected card is a heart or king in 13 + 4 − 1 = 16 ways, because the king of hearts is counted twice otherwise. The probability is 16/52 = 4/13.
For a bag containing 5 red and 3 blue distinct balls, choosing two simultaneously gives 8C2 = 28 equally likely unordered pairs. Two red balls can be chosen in 5C2 = 10 ways, so probability = 10/28 = 5/14. Exactly one red and one blue can be chosen in 5 × 3 = 15 ways, giving 15/28. Keep sample-space conventions consistent: do not count favourable ordered draws while using an unordered denominator.
8. With and without replacement
Replacement changes the experiment. If a ball is returned and the bag is mixed, the second draw again has the original composition. Without replacement, the remaining composition changes. For the bag with 5 red and 3 blue balls, counting ordered pairs without replacement gives 8 × 7 = 56 outcomes and 5 × 4 = 20 red-red outcomes, again 5/14.
With replacement and equally likely individual-ball labels, there are 8 × 8 = 64 ordered outcomes and 5 × 5 = 25 red-red outcomes, giving 25/64. This counting derivation avoids relying on unexplained conditional formulas. State whether replacement occurs before selecting a denominator; the two answers correspond to different random experiments.
9. Case questions and consistency checks
Suppose among equally likely numbered tickets 1 to 30, A is a multiple of 3 and B a multiple of 5. There are 10 multiples of 3, 6 multiples of 5 and 2 multiples of 15. Therefore A or B has 10 + 6 − 2 = 14 tickets, giving 7/15 probability. The overlap is determined by the least common multiple, not by adding the divisors.
Probability data also impose bounds. An intersection cannot exceed either individual probability, and a union cannot be smaller than either. For example, P(A) = 0.8 and P(B) = 0.7 require P(A ∩ B) ≥ 0.5 because the union cannot exceed 1. These checks can reject an inconsistent case-study claim before detailed calculation. Draw a Venn diagram with disjoint regions when several verbal conditions are supplied.
A reliable approach to questions
Define the experiment, outcome order and replacement rules.
Write a sample space or a consistent counting scheme.
Translate exactly, at least, neither, or and both into events.
Choose direct counting, complement or the addition law.
Reduce the fraction and check probability bounds and overlap consistency.
Connections that matter
Sets supplies event unions, intersections, differences and complements.
Permutations and combinations supplies systematic outcome counting.
Statistics uses probability models to interpret variable observations.
Common mistakes
Treating two-dice sums as equally likely ignores different numbers of ordered pairs.
Adding overlapping probabilities without subtraction double-counts outcomes.
Confusing exactly one with at least one includes an unwanted overlap.
Mixing ordered numerators with unordered denominators gives inconsistent counting.
Revision checklist
I can construct a sample space and identify its assumptions.
I can translate verbal conditions into event sets.
I can use complement and addition laws correctly.
I can solve card, dice and selection problems by consistent counting.
I can check probability data for impossible values.