Conditioning and Independence

Abstract

We have seen that the probability of a set A is its weighted proportion relative to the sample space Ω When Ω is finite and all sample points have the same weight (therefore equally likely), then

$$P(A) = \frac{{\left| A \right|}}{{\left| \Omega \right|}}$$

as in Example 4 of §2.2. When Ω is countable and each point ω has the weight P(ω)= P({ω}) attached to it, then

$$P(A) = \frac{{\sum\limits_{{\omega \in A}} {P(\omega )} }}{{\sum\limits_{{\omega \in \Omega }} {P(\omega )} }}$$

(5.1.1)

from (2.4.3), since the denominator above is equal to 1. In many questions we are interested in the proportional weight of one set A relative to another set S. More accurately stated, this means the proportional weight of the part of A in S, namely the intersection A ∩ S, or AS, relative to S. The formula analogous to (5.1.1) is then

$$\frac{{\sum\limits_{\omega \in AS} {P(\omega )} }}{{\sum\limits_{\omega \in S} {P(\omega )} }}.$$

(5.1.2)