Question

Events $A, B, C$ satisfy $P(A)=0.3, P(B)=0.3, P(C)=0.5$. Events $A$ and $B$ are mutually exclusive. Events $A$ and $C$ are independent and $P(B / C)=0.2$. $P(A \cup B \cup C)$ equals

• A. 0.95
• B. 0.85
• C. 0.75
• D. 0.65

Answer 1

Answer: (B)

$P(B / C)=\frac{P(B \cap C)}{P(C)} $
$\therefore P(B \cap C)=0.2 P(C)=0.1$
$\text { now, } P ( A \cup B \cup C )=\sum P ( A )-\sum P ( A \cap B )+ P ( A \cap B \cap C ) $
$=1.1-[0+0.1+(0.5)(0.3)]+0 $
$=1.1-0.25$
$=0.85 \Rightarrow \text { (B) }$