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 ∪ B ∪ C) equals

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 ∪ B ∪ C) equals (A) 0.95 (B) 0.85 (C) 0.75 (D) 0.65

Tardigrade Admin · Accepted Answer

P(B / C)=(P(B ∩ C)/P(C)) ∴ P(B ∩ C)=0.2 P(C)=0.1 text now, P ( A ∪ B ∪ C )=∑ P ( A )-∑ P ( A ∩ B )+ P ( A ∩ B ∩ C ) =1.1-[0+0.1+(0.5)(0.3)]+0 =1.1-0.25 =0.85 ⇒ text (B)

Q. 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

0.95

0.85

0.75

0.65

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

Q. 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∪B∪C) equals