# Conditioning and Independence

Chapter

## Abstract

We have seen that the probability of a set as in Example 4 of §2.2. When Ω is countable and each point 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*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$$P(A) = \left| {\frac{A}{\Omega }} \right|$$

*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)

*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)

## Keywords

Black Ball Conditional Probability Independent Random Variable Conditional Proba Male Black Student
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© Springer Science+Business Media New York 1974