Question

If $A,B,C$ are three events associated with a random experiment, prove that $P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$

Answer 1

Simplification the given data
Consider $B\cup C=E$
$P(A\cup B\cup C)=P(A\cup E)$
As we know that
$P(A\cup E)=P\left(A\right)+P\left(E\right)-P(A\cap E)$
Now,
$P(A\cup B\cup C)=P(A\cup E)$
$\Rightarrow P(A\cup B\cup C)=P\left(A\right)+P\left(E\right)-P(A\cap E)$

$\Rightarrow P(A\cup B\cup C)=P\left(A\right)+P\left(E\right)-P(A\cap (B\cup C\left)\right)$

$\Rightarrow P(A\cup B\cup C)=P\left(A\right)+P\left(E\right)-P\left(\right(A\cap B)\cup (A\cap C\left)\right).....\left(i\right)$

We need to find $P\left(E\right)\&P\left(\right(A\cap B)\cup (A\cap C\left)\right)$

Finding $P\left(E\right)\text{and}P(A\cap B)\cup (A\cap C))$
$P\left(E\right)=P(B\cup C)=P\left(B\right)+P\left(C\right)-P(B\cap C)........\left(ii\right)$

We know that
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

Putting $A=(A\cap B)\&B=(A\cap C)$ then we get,
$P\left(\right(A\cap B)\cup (A\cap C\left)\right)=P(A\cap B)+P(A\cap C)-P\left(\right(A\cap B)\cap (A\cap C\left)\right)$

$P\left(\right(A\cap B)\cup (A\cap C\left)\right)=P(A\cap B)+P(A\cap C)-P(A\cap B\cap C)....\left(iii\right)$

Putting $\left(ii\right)$ & $\left(iii\right)$ in $\left(i\right)$ then we get,

$P(A\cup B\cup C)=P\left(A\right)+\left[P\right(B)+P(C)-P(B\cap C\left)\right]-\left[P\right(A\cap B)+P(A\cap C)-P(A\cap B\cap C\left)\right]$

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