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)

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