Let A- B- C be three events in a probability space- Suppose that PA-0-5- PB-0-3- PC-0-2- P A - B - 0-15-P A - C - 0-1 and P B - C - 0-06The maximum possible value of P A-B-C- is-

The correct option is C 0.31 P(AUBUC)=P(A) + P(B) + P(C) #10; #8722; P(A #8745; B) #8722; P(A #8745; C) #8722; P(B #8745; C) #10;+ P( ...

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Intersection of Sets

Let A, B, C b...

Question

Let A, B, C be three events in a probability space. Suppose that P(A)=0.5, P(B)=0.3, P(C)=0.2,
$P (A \cap B) = 0.15, P (A \cap C) = 0.1$ and $P (B \cap C) = 0.06$
The maximum possible value of $P (A^{'} \cap B^{'} \cap C^{'})$ is,

0.31

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0.25

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Solution

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Similar questions

P (A) = 0.5, P (B) = 0.3, P (C) = 0.2

P (A \cap B) = 0.15, P (A \cap C) = 0.1

P (B \cap C) = 0.06

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

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 \cup B \cup C) \geq 0.85, P (A \cap B \cap C) = 0.2

P (B \cap C) = p_{1}

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

P (B \cap C)

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