If A- B- C are three events- then show that PA - B - C - PA -PB - PC - P A - B - PB - C - P C - A - P A - B - C

We know that P(A ∪ B) = P(A) + P(B) P(A ∩ B) Let us take B ∪ C to be another event R ∴ P(A ∪ B ∪ C) = P(A ∪ R) = P(A) + P(R) P(A ∩ ...

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Addition Theorem of Probability for 2 Events

If A, B, C ...

Question

If $A, B, C$ are three events, then show that $P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)$

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Solution

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Similar questions

A, B, C

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)

P (A) = \frac{1}{4}

P (B) = \frac{1}{6}

P (C) = \frac{2}{3}

P (A \cup B \cup C) = P (A) + P (B) + P (C)

A, B

C

P (A) = 0.3, P (B) = 0.4, P (C) = 0.8, P (A \cup B) = 0.08, P (A \cap C) = 0.28, P (A \cap B \cap C) = 0.09

P (A \cup B \cup C) \geq 0.75

P (B \cap C)

A, B

C

P (A) = 0.3, P (B) = 0.4, P (C) = 0.8

P (A \cap B) = 0.12, P (A \cap C) = 0.28

P (A \cap B \cap C) = 0.09

P (A \cup B \cup C) \geq 0.75

P (B \cap C)

A, B, C

P (A) = 0.3, P (B) = 0.4, P (C) = 0.8, P (A \cap B) = 0.08,

P (A \cap C) = 0.28, P (A \cap B \cap C) = 0.09

P (A \cup B \cup C) \geq 0.75

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