The probabilities of three events A, B and c are given by P(A)=0.6, P(B)=0.4 ,P(C)=0.5 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

∵P(A∩B)=P(A)+P(B)−P(A∪B)=1−0.8=0.2 Now P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)-P(B∩C)-P(C∩A)+P(A∩B∩C)⇒β=1.2−α⇒β∈0.25,0.4

The probabilities of three events A, B and c are given by P(A)=0.6, P(B)=0.4 ,P(C)=0.5 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

∵P(A∩B)=P(A)+P(B)−P(A∪B)=1−0.8=0.2 Now P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)-P(B∩C)-P(C∩A)+P(A∩B∩C)⇒β=1.2−α⇒β∈0.25,0.4

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The probabilities of three events A, B and c are given by P(A)=0.6, P(B)=0.4 ,P(C)=0.5 $P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cap B \cap C) = 0.2, P (B \cap C) = β$ and $P (A \cup B \cup C) = α$ where $0. 85 \leq α \leq 0. 95$ , then $β$ lies in the interval

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[0.35, 0.36]

[0.36, 0.40]

[0.25, 0.35]

[0.20, 0.25]

answer is B.

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Detailed Solution

$\begin{array}{l} ∵ P (A \cap B) = P (A) + P (B) - P (A \cup B) = 1 - 0.8 = 0.2 \\ N o w P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C) \end{array}$
$\begin{array}{l} \Rightarrow β = 1.2 - α \\ \Rightarrow β \in [0.25, 0.4] \end{array}$