Conditional Probability

Motivating conditional probability. Roll a fair die once, $S=\{1,2,3,4,5,6\}$. Now ask two related but different questions.

Q1: What is the probability of getting an odd number which is greater than 2?

Q2: IF the die shows an odd number, then what is the probability that it is greater than 2?

In Case 1 (Q1), the event 'odd number greater than 2' is $\{3,5\}$, so $P_1=\dfrac{n(\{3,5\})}{n(\{1,2,3,4,5,6\})}=\dfrac{2}{6}=\dfrac13$.

In Case 2 (Q2), the phrase 'IF the die shows an odd number' RESTRICTS the sample space itself down to $S_1=\{1,3,5\}$ -- we are only interested in what happens once we already know the outcome is odd. Within that restricted space, the same favourable outcomes $\{3,5\}$ now sit over a smaller total: $P_2=\dfrac{n(\{3,5\})}{n(\{1,3,5\})}=\dfrac{2}{3}$.

The favourable event is identical in both cases, but the exhaustive count of possibilities is different -- because in Case 2 a CONDITION was imposed on the sample space before the probability was computed. This is exactly what conditional probability means. Importantly, the underlying sample space $S$ does not itself change between the two cases; what changes is which count of outcomes the favourable event is being measured against. Written using set notation over the ORIGINAL $S$: $P_2=\dfrac{n(\{3,5\})/n(S)}{n(\{1,3,5\})/n(S)}=\dfrac{P(\{3,5\})}{P(\{1,3,5\})}$, which is exactly $\dfrac{P(A\cap B)}{P(A)}$ for $A=\{1,3,5\}$ (odd) and $B=\{3,5\}$ (greater than 2, i.e. the event 'greater than 2' intersected with 'odd').

Definition 12.14. The conditional probability of an event $B$, ASSUMING that the event $A$ has already happened, is denoted $P(B/A)$ and defined as

Similarly, $P(A/B)=\dfrac{P(A\cap B)}{P(B)}$, provided $P(B)\ne 0$.

Note 12.5. $P(B/A)+P(\bar B/A)=1$ -- given that $A$ has happened, $B$ either happens or it doesn't, so the two conditional probabilities are complementary within the restricted space.

Theorem 12.7 -- the Multiplication Theorem on Probability. Rewriting Definition 12.14 to isolate $P(A\cap B)$ gives the probability of the SIMULTANEOUS happening of two events:

$P(A\cap B)=P(A/B)\,P(B)\qquad\text{or equivalently}\qquad P(A\cap B)=P(B/A)\,P(A).$ …