Question

Events $A,$ $B$ and $C$ satisfy $P\left(A\right)=0.3,$ $P\left(B\right)=0.3,$ $P\left(C\right)=0.5$. If events $A$ and $B$ are mutually exclusive, events $A$ and $C$ are independent, and $P\left(B|C\right)=0.2$, then $300P(A\cup B\cup C)$ equals

• A. 255
• B. 255.00
• C. 255.0

Answer 1

Answer: (A), (B), (C)

$P\left(B|C\right)=0.2,P\left(A\right)=0.3,P\left(B\right)=0.3,P\left(C\right)=0.5$
$P\left(B|C\right)=\frac{P(B\cap C)}{P\left(C\right)}$
$\Rightarrow P(B\cap C)=P\left(B|C\right)\cdot P\left(C\right)=0.2\times 0.5=0.1$
Event $A$ and $C$ are independent.
$\therefore P(A\cap C)=P\left(A\right)\times P\left(C\right)=0.3\times 0.5=0.15$
Events $A$ and $B$ are mutually exclusive.
$\therefore P(A\cap B)=0\Rightarrow P(A\cap B\cap C)=0$

$P(A\cup B\cup C)$
$=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(B\cap C)-P(A\cap C)$
$=0.3+0.3+0.5-0.1-0.15=0.85$