Once outcomes and events are in place, we attach a number — the probability — to each event, measuring how likely it is. The axiomatic approach, due to the Russian mathematician A. N. Kolmogorov, fixes a few simple rules that any sensible probability must obey, and everything else in the chapter follows from them.
For a sample space $S$ and any event $E$, the axioms of probability are:
$$\textbf{(1)}\ \ 0 \le P(E) \le 1 \qquad \textbf{(2)}\ \ P(S) = 1 \qquad \textbf{(3)}\ \ P(A \cup B) = P(A) + P(B) \text{ if } A \cap B = \varnothing$$
In words: (1) every probability lies between $0$ and $1$, (2) the sure event has probability $1$, and (3) for mutually exclusive events the probabilities simply add. From these the impossible event follows immediately: $P(\varnothing) = 0$.
A clean way to assign probabilities is to give each elementary event $\{\omega_i\}$ a number $p_i$ with $0 \le p_i \le 1$ and $p_1 + p_2 + \dots + p_n = 1$; the probability of any event is then the sum of the $p_i$ of the outcomes it contains. This single requirement — that the probabilities of all outcomes add to $1$ — is the practical heart of the axioms.
When all outcomes are equally likely — a fair coin, a fair die, a well-shuffled pack — each $p_i = \dfrac{1}{n(S)}$, and the probability of an event reduces to a ratio of counts. This is the classical (Laplace) definition you will use most:
$$P(A) = \dfrac{n(A)}{n(S)} = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$
Notice that this formula only applies when the outcomes are equally likely. A loaded die or a bent coin breaks the assumption, and then you must use observed (empirical) frequencies instead.
A constant companion of any event is its complement $\overline{A}$, the event "$A$ does not occur". Since $A$ and $\overline{A}$ are mutually exclusive and together exhaust $S$, axiom (3) with $A \cup \overline{A} = S$ gives:
$$P(A) + P(\overline{A}) = 1 \quad\Longrightarrow\quad P(\overline{A}) = 1 - P(A)$$
This complement rule is often the fastest route to an answer: when "at least one" appears in a question, computing $1 - P(\text{none})$ is usually far shorter than counting the favourable cases directly.
Two further consequences of the axioms are worth stating. For any two events (not necessarily disjoint) the probability of "$A$ or $B$" is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, the subtraction removing the doubly-counted overlap (this is the addition theorem, developed fully in the next topic). And monotonicity: if $A \subseteq B$ then $P(A) \le P(B)$ — a larger event is at least as likely.
Odds. Examiners sometimes phrase likelihood as odds. If an event has $m$ favourable and $k$ unfavourable equally likely outcomes, the odds in favour are $m : k$ and the odds against are $k : m$. Converting back, $P(A) = \dfrac{m}{m+k}$. For instance, odds in favour of $3 : 2$ mean $P(A) = \dfrac{3}{5}$.
Deeper Insight — why axioms beat a single "definition": Earlier you may have met probability as nothing more than favourable over total. That classical formula is genuinely useful, but it quietly assumes the outcomes are equally likely — an assumption that fails for a weighted coin, an unfair die, or a continuous quantity like a person's exact height. The axiomatic framework sidesteps this by not committing to how the numbers arise; it only demands that whatever numbers we assign are non-negative, sum to $1$ over the whole sample space, and add up over disjoint events. The familiar $\tfrac{n(A)}{n(S)}$ is then just one way to satisfy the axioms — the way that happens to work when symmetry makes every outcome equally likely. Because the complement rule, the addition theorem, monotonicity, and every later result are all derived from these axioms, learning to reason from them — rather than memorising isolated formulas — is what makes harder problems tractable.
A fair die is rolled. Find the probability of getting (a) a number greater than $4$, (b) a multiple of $3$.
Solution- $n(S) = 6$ equally likely outcomes.
- (a) Greater than $4$: $\{5, 6\}$, so $P = \dfrac{2}{6} = \dfrac{1}{3}$.
- (b) Multiples of $3$: $\{3, 6\}$, so $P = \dfrac{2}{6} = \dfrac{1}{3}$.
Answer: (a) $\dfrac{1}{3}$; (b) $\dfrac{1}{3}$.
A card is drawn from a well-shuffled pack of $52$. Find the probability that it is (a) a king, (b) a face card.
Solution- $n(S) = 52$ equally likely cards.
- (a) There are $4$ kings: $P = \dfrac{4}{52} = \dfrac{1}{13}$.
- (b) Face cards are J, Q, K in each suit: $3 \times 4 = 12$, so $P = \dfrac{12}{52} = \dfrac{3}{13}$.
Answer: (a) $\dfrac{1}{13}$; (b) $\dfrac{3}{13}$.
If $P(A) = \dfrac{2}{5}$, find $P(\overline{A})$.
Solution- Use the complement rule: $P(\overline{A}) = 1 - P(A)$.
- $P(\overline{A}) = 1 - \dfrac{2}{5} = \dfrac{3}{5}$.
Answer: $P(\overline{A}) = \dfrac{3}{5}$.
Three coins are tossed together. Find the probability of getting at least one tail.
Solution- $n(S) = 2^3 = 8$ equally likely outcomes.
- "At least one tail" is the complement of "no tail" (all heads, the single outcome $HHH$).
- $P(\text{no tail}) = \dfrac{1}{8}$, so $P(\text{at least one tail}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$.
Answer: $\dfrac{7}{8}$.
Two dice are thrown. Find the probability that the sum of the numbers is $8$.
Solution- $n(S) = 36$ equally likely ordered pairs.
- Pairs giving a sum of $8$: $(2,6),(3,5),(4,4),(5,3),(6,2)$ — that is $5$ favourable outcomes.
- $P = \dfrac{5}{36}$.
Answer: $\dfrac{5}{36}$.
A bag holds $4$ red and $6$ blue balls. One ball is drawn at random. Find the probability it is (a) red, (b) not red.
Solution- Total balls $n(S) = 4 + 6 = 10$, each equally likely to be drawn.
- (a) Red: $P(\text{red}) = \dfrac{4}{10} = \dfrac{2}{5}$.
- (b) By the complement rule, $P(\text{not red}) = 1 - \dfrac{2}{5} = \dfrac{3}{5}$ (check: $\dfrac{6}{10} = \dfrac{3}{5}$).
Answer: (a) $\dfrac{2}{5}$; (b) $\dfrac{3}{5}$.
A die is loaded so that each even face is twice as likely as each odd face. Find the probability of each outcome and hence $P(\text{even})$.
Solution- Let an odd face have probability $p$; then each even face has $2p$.
- Axiom: probabilities sum to $1$, so $3p + 3(2p) = 9p = 1 \Rightarrow p = \dfrac{1}{9}$.
- Odd faces: $\dfrac{1}{9}$ each; even faces: $\dfrac{2}{9}$ each. $P(\text{even}) = 3 \times \dfrac{2}{9} = \dfrac{6}{9} = \dfrac{2}{3}$.
Answer: Odd faces $\dfrac{1}{9}$, even faces $\dfrac{2}{9}$; $P(\text{even}) = \dfrac{2}{3}$. (The classical $\tfrac{n(A)}{n(S)}$ would not apply here.)
The probability that it rains tomorrow is $0.35$. Find the probability that it does not rain. If the odds in favour of an event are $5 : 3$, find its probability.
Solution- Complement rule: $P(\text{no rain}) = 1 - 0.35 = 0.65$.
- Odds in favour $5 : 3$ mean $5$ favourable to $3$ unfavourable, total $8$ parts.
- $P = \dfrac{5}{5+3} = \dfrac{5}{8}$.
Answer: $P(\text{no rain}) = 0.65$; the event with odds $5:3$ has probability $\dfrac{5}{8}$.
Check whether each is a valid assignment of probabilities to the outcomes $\omega_1, \omega_2, \omega_3$: (a) $0.3, 0.4, 0.3$; (b) $0.5, 0.6, -0.1$; (c) $\tfrac14, \tfrac14, \tfrac14$.
Solution- Two tests: every value in $[0,1]$, and the values sum to $1$.
- (a) All in range and $0.3+0.4+0.3 = 1$ — valid.
- (b) $-0.1 < 0$ — fails axiom (1), invalid.
- (c) $\tfrac14 \times 3 = \tfrac34 \ne 1$ — the total is not $1$, invalid.
Answer: Only (a) is a valid probability assignment.
A number is chosen at random from the first $20$ natural numbers. Find the probability that it is (a) a multiple of $4$, (b) a prime, (c) not a prime.
Solution- $n(S) = 20$ equally likely numbers $1$ to $20$.
- (a) Multiples of $4$: $4,8,12,16,20$ — $5$ of them, $P = \dfrac{5}{20} = \dfrac{1}{4}$.
- (b) Primes $\le 20$: $2,3,5,7,11,13,17,19$ — $8$ of them, $P = \dfrac{8}{20} = \dfrac{2}{5}$.
- (c) Complement: $P(\text{not prime}) = 1 - \dfrac{2}{5} = \dfrac{3}{5}$.
Answer: (a) $\dfrac{1}{4}$; (b) $\dfrac{2}{5}$; (c) $\dfrac{3}{5}$.
Two dice are thrown. Find the probability that (a) both show the same number, (b) the numbers differ.
Solution- $n(S) = 36$. Doublets $(1,1),\dots,(6,6)$ number $6$.
- (a) $P(\text{same}) = \dfrac{6}{36} = \dfrac{1}{6}$.
- (b) "Differ" is the complement: $P = 1 - \dfrac{1}{6} = \dfrac{5}{6}$.
Answer: (a) $\dfrac{1}{6}$; (b) $\dfrac{5}{6}$.
The letters of the word "PROBABILITY" are written on cards and one is drawn at random. Find the probability that it is the letter (a) B, (b) a vowel.
Solution- "PROBABILITY" has $11$ letters, so $n(S) = 11$ equally likely cards.
- (a) The letter B appears twice (in PROBABILITY), so $P(\text{B}) = \dfrac{2}{11}$.
- (b) Vowels are O, A, I, I — that is $4$ cards, so $P(\text{vowel}) = \dfrac{4}{11}$.
Answer: (a) $\dfrac{2}{11}$; (b) $\dfrac{4}{11}$.