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The probabilities of three events A, B and C are given by $P (A) = 0.6, P (B) = 0.4 a n d P (C) = 0.5 .$ 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 the maximum possible value of $β$ can be

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a

0.05

b

0.35

c

0.25

d

0.15

answer is B.

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Detailed Solution

$∵ P (A \cap B) = P (A) + P (B) - P (A \cup B) = 1 - 0.8 = 0.2$

NOW, $∵ P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B)$

$- P (B \cap C) - P (C \cap A) + P (A \cap B \cap C) \Rightarrow α = 0.6 + 0.4 + 0.5 - 0.2 - β - 0.3 + 0.2 \Rightarrow β = 1.2 - α$

$∵ α \in [0.85, 0.95] t h e n β \in [0.25, 0.35]$

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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∪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 the maximum possible value of β can be