Suppose P(A | B) = 1/2 and P(A ? B) = 1/6. What is P(B)?
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Textbook Solutions for Discrete Mathematics with Applications
Question
Prove Bayes’ Theorem for n = 2. That is, prove that if a sample space S is a union of mutually disjoint events \(B_{1}\) and \(B_{2}\), if A is an event in S with P(A) = 0, and if k = 1
or k = 2, then
\(P\left(B_{k} \mid A\right)=\frac{P\left(A \mid B_{k}\right) P\left(B_{k}\right)}{P\left(A \mid B_{1}\right) P\left(B_{1}\right)+P\left(A \mid B_{2}\right) P\left(B_{2}\right)}\).
Text Transcription:
P (B_k | A) = P(A | B_k) P(B_k) / P(A | B_1) P(B_1) + (P(A | B_2) P(B_2)
Solution
The first step in solving 9.9 problem number 9 trying to solve the problem we have to refer to the textbook question: Prove Bayes’ Theorem for n = 2. That is, prove that if a sample space S is a union of mutually disjoint events \(B_{1}\) and \(B_{2}\), if A is an event in S with P(A) = 0, and if k = 1or k = 2, then\(P\left(B_{k} \mid A\right)=\frac{P\left(A \mid B_{k}\right) P\left(B_{k}\right)}{P\left(A \mid B_{1}\right) P\left(B_{1}\right)+P\left(A \mid B_{2}\right) P\left(B_{2}\right)}\).Text Transcription:P (B_k | A) = P(A | B_k) P(B_k) / P(A | B_1) P(B_1) + (P(A | B_2) P(B_2)
From the textbook chapter Conditional Probability, Bayes’ Formula, and Independent Events you will find a few key concepts needed to solve this.
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