Q.Four persons are selected at random from a group of 3 men, 2 women and 4 children. The probability that exactly two of them are children is

Classical (a priori) definition. When a random experiment's outcomes are all equally likely, the classical or a priori probability of an event $A$ is

This is the everyday 'favourable over total' rule, and it silently needs two things: the outcomes must be equally likely, and there must be finitely many of them. Neither the coin-till-first-head experiment (infinite outcomes) nor a biased die (unequal chances) can be handled by this definition alone -- which is why the axiomatic approach was developed.

Axiomatic approach (Kolmogorov, 1933). Let $S$ be a finite sample space, $\mathcal P(S)$ the class of all events, and $P$ a real-valued function on $\mathcal P(S)$. $P(A)$ is a probability function exactly when it obeys three axioms:

• $[P_1]$ Non-negativity: $P(A)\ge 0$ for every event $A$.
• $[P_2]$ Additivity: for mutually exclusive $A,B$: $P(A\cup B)=P(A)+P(B)$ (and more generally, for mutually exclusive $A_1,\ldots,A_n$: $P(A_1\cup\cdots\cup A_n)=P(A_1)+\cdots+P(A_n)$).
• $[P_3]$ Normalisation: $P(S)=1$.

From these, $0\le P(A)\le 1$ always. Theorem 12.1 shows the classical ratio $P(A)=n(A)/n(S)$ automatically satisfies all three axioms -- so classical probability is one particular case of the axiomatic theory, not a rival to it. Theorem 12.2 shows the same for any finite probability space: assign each sample point $a_i$ a real number (probability) $p_i\ge 0$ with $\sum p_i=1$, define $P(A)$ as the sum of the $p_i$ for points inside $A$, and the three axioms hold again -- this is the route to probabilities that are NOT equally likely (Illustration 12.6 gives examples where the $p_i$ differ, and even irrational $p_i$ are allowed as long as they are non-negative and sum to 1).

Odds. If $a$ is the number of ways an event $A$ can occur and $b$ the number of ways it fails, the odds in favour of $A$ are $a:b$, equivalently $P(A)=\dfrac{a}{a+b}$; the odds against $A$ are $b:a$. If $P(A)=p$ is already known, the odds in favour are $p:(1-p)$ and the odds against are $(1-p):p$.