Question

Prove that$P\left(E{F}^c\right)=P\left(E\right)-P\left(EF\right).$

Short answer

Apply Axiom $3$for mutually exclusive events role="math" localid="1649247228736" $EF$and$E{F}^c$.

Step-by-step solution

Given Information.

$P\left(E{F}^c\right)=P\left(E\right)-P\left(EF\right)$

Explanation.

For $E$and $F$events within some outcome space:

$P\left(E{F}^c\right)=P\left(E\right)-P\left(EF\right)$

Events $EF$& localid="1649247667317" $E{F}^c$are mutually exclusive because they $X$would be an element of that intersection $x\in EF\Rightarrow x\in F$and localid="1649247675420" $x\in E{F}^c\Rightarrow x\in F^c$that is a contradiction.

$EF\cup E{F}^c=E$this is the first Theoretical exercise in this section

These two facts lead to the use of the Axiom$3$:

$P\left(EF\cup E{F}^c\right)=P\left(E{F}^c\right)+P\left(EF\right)\Rightarrow P\left(E\right)=P\left(E{F}^c\right)+P\left(EF\right)$

And the right to equality is easily transformed into the wanted identity.