The probabilities of three events A , B and C are given by P ( A )=0.6, P ( B )=0.4 and P ( C )=0.5 If P ( A ∪ B )=0.8, P ( A ∩ C )=0.3, P ( A ∩ B ∩ C ) =0.2, P ( B ∩ C )=β and P ( A ∪ B ∪ C )=α where 0.85 ≤ α ≤ 0.95, then β lies in the interval:

Question

The probabilities of three events A , B and C are given by P ( A )=0.6, P ( B )=0.4 and P ( C )=0.5 If P ( A ∪ B )=0.8, P ( A ∩ C )=0.3, P ( A ∩ B ∩ C ) =0.2, P ( B ∩ C )=β and P ( A ∪ B ∪ C )=α where 0.85 ≤ α ≤ 0.95, then β lies in the interval: (A) [0.36, 0.40] (B) [0.35, 0.36] (C) [0.25, 0.35] (D) [0.20, 0.25]

Tardigrade Admin · Accepted Answer

P ( A ∪ B )= P ( A )+ P ( B )- P ( A ∩ B ) 0.8=0.6+0.4- P ( A ∩ B ) P ( A ∩ B )=0.2 P ( A ∪ B ∪ C )= Sigma P ( A )- Sigma P ( A ∩ B )+ P ( A ∩ B ∩ C ) α=1.5-(0.2+0.3+β)+0.2 α=1.2-β ∈[0.85,0.95] (where α ∈[0.85,0.95]) β ∈[0.25,0.35]

Q. The probabilities of three events $A, B$ and $C$ are given by $P (A) = 0.6, P (B) = 0.4$ and $P (C) = 0.5$ If $P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cap B \cap C)$ $= 0.2, P (B \cap C) = β$ and $P (A \cup B \cup C) = α$ where $0.85 \leq α \leq 0.95,$ then $β$ lies in the interval:

$[0.36, 0.40]$

$[0.35, 0.36]$

$[0.25, 0.35]$

$[0.20, 0.25]$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$0.8 = 0.6 + 0.4 - P (A \cap B)$
$P (A \cap B) = 0.2$
$P (A \cup B \cup C) = Σ P (A) - Σ P (A \cap B) + P (A \cap B \cap C)$
$α = 1.5 - (0.2 + 0.3 + β) + 0.2$
$α = 1.2 - β \in [0.85, 0.95]$
(where $α \in [0.85, 0.95])$
$β \in [0.25, 0.35]$

Q. The probabilities of three events A,B and C are given by P(A)=0.6,P(B)=0.4 and P(C)=0.5 If P(A∪B)=0.8,P(A∩C)=0.3,P(A∩B∩C) =0.2,P(B∩C)=β and P(A∪B∪C)=α where 0.85≤α≤0.95, then β lies in the interval:

[0.36,0.40]

[0.35,0.36]

[0.25,0.35]

[0.20,0.25]