Let $A, B$ and $C$ be three events associated with sample spaces $S . A, B$ and $C$ are pair wise…
Let $A, B$ and $C$ be three events associated with sample spaces $S . A, B$ and $C$ are pair wise independent and $P(A)=P(B)=P(C)=P$. If all of them cannot occur simultaneously, then $P(A \cup B \cup C)$ is equal to
$1-(1-P)^3$
$3 P(1-P)$
$P^3$
$3 P$
Solution
A, B, C are pair-wise independent events.
$
\begin{aligned}
\Rightarrow \quad & P(A \cap B)=P(A) \cdot P(B) \\
& P(B \cap C)=P(B) \cdot P(C) \\
& P(C \cap A)=P(C) \cdot P(A)
\end{aligned}
$
and $P(A)=P(B)=P(C)=P$
and $P(A \cap B \cap C)=0$
[ $\because$ they cannot occur simultaneously]
Now, $P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)$
$-P(B \cap C)-P(C \cap A)+P(A \cap B \cap C)$
$=P+P+P-P(A) \cdot P(B)-P(B) \cdot P(C)-P(C) \cdot P(A)+0$
$=3 P-P \cdot P-P \cdot P-P \cdot P=3 P-P^2-P^2-P^2$
$=3 P-3 P^2=3 P(1-P)$