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*2.100 Show that Theorem 2.6, the additive law of probability, holds for conditional probabilities. That is, if A, B, and C are events such that P(C) > 0, prove that P(AU B|C) = P(A|C) + P(B|C)–P(ANB|C). [Hint: Make use of the distributive law (AUB)NC = (ANC)U(BNC).] The Additive Law of Probability The probability of the union of two events A and B is THEOREM 2.6 P(A U B) = P(A) + P(B) – P(AN B). If A and B are mutually exclusive events, P(A N B) = 0 and P(AU B) = P(A)+ P(B).

*2.100 Show that Theorem 2.6, the additive law of probability, holds for conditional probabilities. That is, if A, B, and C are events such that P(C) > 0, prove that P(AU B|C) = P(A|C) + P(B|C)–P(ANB|C). [Hint: Make use of the distributive law (AUB)NC = (ANC)U(BNC).] The Additive Law of Probability The probability of the union of two events A and B is THEOREM 2.6 P(A U B) = P(A) + P(B) – P(AN B). If A and B are mutually exclusive events, P(A N B) = 0 and P(AU B) = P(A)+ P(B).

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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*2.100 Show that Theorem 2.6, the additive law of probability, holds for conditional probabilities. That is, if A, B, and C are events such that P(C) > 0, prove that P(A U B|C) = P(A|C) + P(B|C)–P(ANB|C). [Hint: Make use of the distributive law (AUB)NC = (ANC)U(BNC).] The Additive Law of Probability The probability of the union of two events A and B is THEOREM 2.6 P(AUB) = P(A) + P(B) – P(AN B). If A and B are mutually exclusive events, P(AN B) = 0 and P(AU B) = P(A)+ P(B).
Transcribed Image Text:*2.100 Show that Theorem 2.6, the additive law of probability, holds for conditional probabilities. That is, if A, B, and C are events such that P(C) > 0, prove that P(A U B|C) = P(A|C) + P(B|C)–P(ANB|C). [Hint: Make use of the distributive law (AUB)NC = (ANC)U(BNC).] The Additive Law of Probability The probability of the union of two events A and B is THEOREM 2.6 P(AUB) = P(A) + P(B) – P(AN B). If A and B are mutually exclusive events, P(AN B) = 0 and P(AU B) = P(A)+ P(B).
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Given,

A, B and C are events such that P(C)>0,

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To show

P(A∪B/C) = P(A/C) + P(B/C) -P(A∩B/C)

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