Prove that $P(E\cup F\cup G)=P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(E^{c}FG\right)-P\left(E{F}^{c}G\right)-P\left(EF{G}^c\right)-2P\left(EFG\right)$

To prove:

$$\begin{gathered} P(E\cup F\cup G)=P\left(E\right)+P\left(F\right)+P\left(G\right)- \\ P\left(E^{c}FG\right)-P\left(E{F}^{c}G\right)-P\left(EF{G}^c\right)-2P\left(EFG\right) \end{gathered}$$

Proof will be similar to the proof of Proposition 4.4. in the remark.

If the outcome from the outcome space is not in $E$or $F$or $G$then its probability does not contribute to $P(E\cup F\cup G)$nor to any probability on the left.

If the outcome from the outcome space is in $E\cup F\cup G$it is in either one, two or all three events $E,F$or $G$. In each of these cases it is counted once in the left hand side.

If it is in precisely one of the events $E,F,G$it is counted only once and that is in the one of the first three addends on the right hand side

If it is in precisely two of the events $E,F,G$it is counted twice in the first three addends, and -l times in one of the second three addends, and $0$times$P(EFG)$. That is it is counted once on the right-hand side.

If it is in all three of the events $E,F,G$ it is counted three times in the first three addends on the right hand side, zero times in the second three addends (because it is not not a member of any of the events) and $-2$ times in $P(EFG)$, so precisely once on the right hand side.

This proves the statement.

$EFG$and $EF\left(G^c\right)$are mutualy exclusive and $EF=EFG\cup EF{G}^c$, so from Axiom $3$:

$P\left(EF\right)=P\left(EFG\right)+P\left(EF\left(G^c\right)\right)$

Since this is true for any three events:

localid="1650021803779" $$\begin{gathered} P\left(FG\right)=P\left(EFG\right)+P\left(E^{c}FG\right) \\ P\left(EG\right)=P\left(EFG\right)+P\left(E{F}^{c}G\right) \end{gathered}$$

Starting from Proposition 4.4.

localid="1650021814055" $$\begin{gathered} P(E\cup F\cup G)=P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(EF\right)-P\left(EG\right)-P\left(FG\right)+P\left(EFG\right) \\ =P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(EFG\right)-P\left(EF{G}^c\right) \\ -P\left(EFG\right)-P\left(E{F}^{c}G\right)-P\left(EFG\right)-P\left(E^{c}FG\right)+P\left(EFG\right) \\ =P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(EF{G}^c\right)-P\left(E{F}^{c}G\right)-P\left(E^{c}FG\right)-2P\left(EFG\right) \end{gathered}$$

It proves the statement.