Part 1: Suppose that F and X are events from a common sample space with P(F) +0 and P(X)+0. P(X|F)P(F) + P(X|F)P(F). Hint: Explain why P(X|F)P(F) (a) Prove that P(X) P(X N F) is another way of writing the definition of conditional probability, and then use that with the logic from the proof of Theorem 4.1.1. %3D (b) Explain why P(F|X) = P(X|F)P(F)/P(X) is another way of stating Theorem 4.2.1 Bayes’ Theorem.

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Chapter1: Combinatorial Analysis
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I am needing assistance with both parts a and b.

***Please type your answer or write in print, because I have had great difficulty with understanding most handwritten assistance done in cursive or mixed print/cursive.

Part 1: Suppose that F and X are events from a common sample space with P(F) +0 and P(X)+0.
(a) Prove that P(X) = P(X|F)P(F)+ P(X|F)P(F). Hint: Explain why P(X|F)P(F)
P(X N F) is another way of writing the definition of conditional probability, and then use
that with the logic from the proof of Theorem 4.1.1.
(b) Explain why P(F|X) = P(X|F)P(F)/P(X) is another way of stating Theorem 4.2.1 Bayes'
Theorem.
Transcribed Image Text:Part 1: Suppose that F and X are events from a common sample space with P(F) +0 and P(X)+0. (a) Prove that P(X) = P(X|F)P(F)+ P(X|F)P(F). Hint: Explain why P(X|F)P(F) P(X N F) is another way of writing the definition of conditional probability, and then use that with the logic from the proof of Theorem 4.1.1. (b) Explain why P(F|X) = P(X|F)P(F)/P(X) is another way of stating Theorem 4.2.1 Bayes' Theorem.
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