The probabilities of three events A-B and C are P A -0-6-P B -0-4-P C -0-5-also P A- B -0-8-P A- C -0-3-P A- B- C - 0-85-P A- B- C -0-2 and P B- C - p 1- Then

The correct option is A 0.2≤ p _ 1 ≤ 0.35 0.85≤ P( A∪ B∪ C ) ≤ 1 P( A∩ B ) =P( A ) +P( B ) P( A∪ B ) =0.2 ∴ P( A∪ B∪ C ) =0.6+0.4+0.5 0.2 0 ...

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Intersection of Sets

The probabili...

Question

The probabilities of three events $A, B$ and $C$ are $P (A) = 0.6, P (B) = 0.4, P (C) = 0.5$ ,also
$P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cup B \cup C) \geq 0.85, P (A \cap B \cap C) = 0.2$
and $P (B \cap C) = p_{1}$ . Then

p_{1} \geq 0.35

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p_{1} \leq 0.2

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0.2 \leq p_{1} \leq 0.35

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None of these

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Solution

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Similar questions

A, B

C

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 (A \cup B \cup C) \geq 0.85

0.04 k \leq P (B \cap C) \leq 0.07 k

k

A, B

C

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 (A \cup B \cup C) \geq 0.85

A, B, C

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 (A \cup B \cup C) \geq 0.85

P (B \cap C)

P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cap B \cap C) = 0.2

P (A \cup B \cup C) \geq 0.85

P (B \cap C)

P (A) = 0.3, P (B) = 0.4, P (C) = 0.5, P (A \cap B^{^{'}}) = 0.2, P (B \cap C) = 0.3, P (A^{^{'}} \cap B^{^{'}} \cap C^{^{'}}) = 0.3

P (A \cap B | C^{^{'}}) = 0.1

P (B^{^{'}} | C^{^{'}})

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