Q.10 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

Use Proposition 4.4. and $P\left(AB\right)=P\left(ABC\right)+P\left(A{B}^c\right)$.

Step 1: Given Information

To prove:

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

Step 2: Explanation

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$P(E F G)$. 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 $P(E F G)$, so precisely once on the right hand side.

This proves the statement.

Step 3: Explanation

$E F G$ 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:

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

Starting from Proposition 4.4.

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

It proves the statement.