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Conditional Probability and Bayes Theorem in Discrete Mathematics

snhu MAT 230 EXAM TWO This document is proprietary to Southern New Hampshire University. It and the problems within may not be posted on any non-SNHU website. Sharmaine Manlansing 1 snhu Directions: Type your solutions into this document and be sure to show all steps for arriving at your solution. Just giving a final number may not receive full credit. PROBLEM 1 This question has 2 parts. 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. To prove this: We utilize conditional probability for events F and X, given that P(F) > 0 and P(X) > 0. The conditional probability for X given F is: P(X|F) = P(F) P(XnF) For the complement of F, the conditional probability is: P(X|F)= P(XnF) P(F) To prove the right-hand side: P(X|F).P(F)+P(XF).P(F) Rewrite: = P(X|F).P(F)+P(XF).P(F) _P(XOF).P(F)+P(XnF).P(F) P(F)+P(F) = P(Xn(FUF)), where FnF ¥ 0 = P(Xn), where= (X) This leads to the conclusion: P(X)=P(X|F).P(F)+P(X|F).P(F) (b) Explain why P(F|X)=P(X|F)P(F)/P(X) is another way of stating Theorem 4.2.1 Bayes Theorem. P(X|F)×P(F)=P(XNF) P(X|F)×P(F)=P(X|F)×P(F) Thus, we conclude: P(F(X))=P(X|F)×P(F)+P(X|F)×P(F) Part 2: A website reports that 70% of its users are from outside a certain country. Out of their users from outside the country, 60% of them log on every day. Out of their users from inside the country, 80% of them log on every day. snhu (a) What percent of all users log on every day? Hint: Use the equation from Part 1 (a). A: Out of the country B:Out of the country (every day) C: In the country (everyday) P(A) =. 70 P(C/A) =. 60 P(C/B) =. 80 P(C/A) =. 80 P(B) = P(A) = 1 -. 70 -. 30 P(C)=P(C|A)*P(A)+P(C|?)*P(A) (.60 * . 70) + (.80 * . 30) .42 +.24 P(C) =. 66 (b) Using Bayes Theorem, out of users who log on every day, what is the probability that they are from inside the country? P(B/C)=P(B)*P(C|B)/P(A)*P(C|A)+P(B)*P(C/B) .30 * . 80/.66 P(B|C)=4/11