Solution Found!
Show that Theorem 2.6, the additive law of probability,
Chapter 2, Problem 100E(choose chapter or problem)
Show that Theorem , the additive law of probability, holds for conditional probabilities. That is, if
, and
are events such that \(P(C)>0\), prove that \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\). Hint: Make use of the distributive law \((A \cup B) C=(A \cap C) \cup(B \cap C)\).]
Equation Transcription:
Text Transcription:
P(C)>0
P(AUB|C)=P(A|C)+P(B|C)-P(A cap B|C)
(AUB)C=(A cap C)U(B cap C)
Questions & Answers
QUESTION:
Show that Theorem , the additive law of probability, holds for conditional probabilities. That is, if
, and
are events such that \(P(C)>0\), prove that \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\). Hint: Make use of the distributive law \((A \cup B) C=(A \cap C) \cup(B \cap C)\).]
Equation Transcription:
Text Transcription:
P(C)>0
P(AUB|C)=P(A|C)+P(B|C)-P(A cap B|C)
(AUB)C=(A cap C)U(B cap C)
ANSWER:
Solution :
Step 1 of 1:
Let A, B, and C are three events.
Here if A, b. and C are events such that P(C)>0.
Our goal is:
We need to prove that .