9.2 Fundamentals

101

The general approach to solving a problem requiring probability is as follows:

1. Choose a set to represent the possible outcomes;

2. Allocate probabilities to these possible outcomes.

The results of probability theory can be derived from three basic axioms, referring to

events and their totality in a manner that we must take to be carefully circumscribed: ⁵

StartLayout 1st Row 1st Column Blank 2nd Column upper P left brace upper E right brace 3rd Column greater than or equals 0 for every event upper E comma 2nd Row 1st Column Blank 3rd Row 1st Column Blank 2nd Column upper P left brace upper S right brace 3rd Column equals 1 for the certain event upper S comma 4th Row 1st Column Blank 5th Row 1st Column Blank 2nd Column upper P left brace upper A right brace 3rd Column equals sigma summation Underscript i Endscripts upper P left brace a Subscript i Baseline right brace period EndLayout P{E} ≥0 for every event E ,

(9.1)

P{S} = 1 for the certain event S ,

(9.2)

P{A} =

E

i

P{ai} .

(9.3)

upper SS includes all possible outcomes. Hence, if upper EE and upper FF are mutually exclusive events,

the probability of their joint occurrence (corresponding to the AND relation in logic;

i.e., “upper EE and upper FF”) is simply the sum of their probabilities:

upper P left brace upper E union upper F right brace equals upper P left brace upper E right brace plus upper P left brace upper F right brace periodP{E ∪F} = P{E} + P{F} .

(9.4)

Simple events are by definition mutually exclusive (upper P left brace upper E right brace intersection upper P left brace upper F right brace equals 0P{E} ∩P{F} = 0), but com-

pound events may include some simple events that belong to other compound events

and, more generally (inclusive OR; i.e., “upper EE or upper FF or both”),

upper P left brace upper E union upper F right brace equals upper P left brace upper E right brace plus upper P left brace upper F right brace minus upper P left brace upper E upper F right brace periodP{E ∪F} = P{E} + P{F} −P{E F} .

(9.5)

If events are independent, then the probability of occurrence of those portions shared

by both is

upper P left brace upper E intersection upper F right brace equals upper P left brace upper E upper F right brace equals upper P left brace upper E right brace upper P left brace upper F right brace periodP{E ∩F} = P{E F} = P{E}P{F} .

(9.6)

It follows that for equally likely outcomes (such as the possible results from throwing

a die or selecting from a pack of cards), the probabilities of compound events are

proportional to the number of equally probable simple events that they contain:

upper P left brace upper A right brace equals StartFraction upper N left brace upper A right brace Over upper N left brace upper S right brace EndFraction periodP{A} = ^N{A}

N{S} ^.

(9.7)

We used this result at the beginning of this section to deduce that the probability of

obtaining a 5 from the throw of a die is 1/6.

Problem. Prove Eqs. (9.4) and (9.5) with the help of Venn diagrams.

5 Notation: in this chapter,upper P left brace upper X right braceP{X} denotes the probability of eventupper XX;upper N left brace upper X right braceN{X} is the number of simple

events in (compound) eventupper XX.upper SS denotes a certain event that contains all possible events. Sample

space and events are primitive (undefined) notions (cf. line and point in geometry).