Show that$\mathit{P}\left(A+B+C\right)\mathbf{=}\mathit{P}\left(A\right)\mathbf{+}\mathit{P}\left(B\right)\mathbf{+}\mathit{P}\left(C\right)\mathbf{-}\mathit{P}\left(AB\right)\mathbf{-}\mathit{P}\left(AC\right)\mathbf{-}\mathit{P}\left(BC\right)\mathbf{+}\mathit{P}\left(ABC\right)$.

Hint: Start with Figure 3.2 and sketch in a region C overlapping some of the pointsof each of the regions A, B, and AB.

$A,B$and $C$are the events with probabilities $P\left(A\right),P\left(B\right)$and $P\left(C\right)$respectively.

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

When events are independent, apply the formula $P\left(AB\right)=P\left(A\right).P\left(B\right)$here$A$and are the events.

Draw the regions for A , B and C.

Take the left hand side $P\left(A+B+C\right)$as $P\left(A+\left(B+C\right)\right)$and apply the formula$P\left(A+B\right)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$.

$P\left(A+\left(B+C\right)\right)=P\left(A\right)+P\left(B+C\right)-P\left(A\left(B+C\right)\right)$

Apply the distributive law.

$P\left(A+\left(B+C\right)\right)=P\left(A\right)+P\left(B+C\right)-P\left(A\left(B+C\right)\right)$

Again, use the formula $P\left(A+B\right)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$and simplify.

$$\begin{gathered} P\left(A+\left(B+C\right)\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(BC\right)-P\left(AB+AC\right) \\ =P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(BC\right)-P\left(AB\right)-P\left(AC\right)+P\left(ABC\right) \end{gathered}$$

Apply the associative law to solve.

$P\left(A+\left(B+C\right)\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(BC\right)-P\left(AB\right)-P\left(AC\right)+P\left(ABC\right)$

It can be observed that right side of the obtained equation is same as the right side of the given equation, thus the given equation is valid.