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 :

P(A∪B∪C)=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(C∩A)+P(A∩B∩C)⇒α=1.2-β0.85≤α≤0.95⇒0.85≤1.2-β≤0.95 ∣⇒0.25≤β≤0.35

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 :

P(A∪B∪C)=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(C∩A)+P(A∩B∩C)⇒α=1.2-β0.85≤α≤0.95⇒0.85≤1.2-β≤0.95 ∣⇒0.25≤β≤0.35

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

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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

$\begin{array}{l} P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) \\ - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C) \\ \Rightarrow α = 1.2 - β \\ 0.85 \leq α \leq 0.95 \\ \Rightarrow 0.85 \leq 1.2 - β \leq 0.95 ∣ \Rightarrow 0.25 \leq β \leq 0.35 \end{array}$