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Conditional Probability and Hasse Diagrams

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. Jamar Sampson 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. P(X NF)=P(X|F)/P(F)isatruestatement. This reasoning is from gained from getting all probabilities of both above listed events while avoiding counting twice events that have already occurred. P(X NF)isequaltoP(XNF)*P(F). All probabilities must add up to one and not go over. Therefore all listed values equal to 1, backing up the initial statement. (b) Explain why P(F|X)=P(X|F)P(F)/P(X) is another way of stating Theorem 4.2.1 Bayes Theorem. Let us prove that P(F(X)) = P(x-F) *P(F) - P(X) using Bayer's Theorem. P(X|F)=P(XnF)/P(F) P(XNF)=P(X|F)*P(F) P(X|F)+P(FnX)/P(X) Since P(F NX)equalsP(X NF) P(X |F)equalsP(X NF)dividedbyP(X). The end result is P(FnX)=P(X|F)*P(F)/P(X)forall. 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. (a) What percent of all users log on every day? Hint: Use the equation from Part 1 (a). O = Foreign Users I = Domestic Users L = Everyday Users P(O) = 0.7 P(I) = 0.8 P(E) = 0.6 P(X)=P(X|F)*P(F)+P(X|F)*P(F) This means (0.6 * 0.7) + (0.8 * 0.3) = 0.66. Our percentage is 0.66. (b) Using Bayes Theorem, out of users who log on every day, what is the probability that they are from inside the country? P(I|L)=(0.3*0.8)dividedby(0.66). This is also can be represented as (4 / 11). This represents the probability that the users are from inside the country. snhu PROBLEM 2 This question has 2 parts. Part 1: The drawing below shows a Hasse diagram for a partial order on the set: {A, B, C, D, E, F, G, H, I, J} D G J C H E B I F A Figure 1: A Hasse diagram shows 10 vertices and 8 edges. The vertices, represented by dots, are as follows: vertex J is upward of vertex H; vertex H is upward of vertex I; vertex B is inclined upward to the left of vertex A; vertex C is upward of vertex B; vertex D is inclined upward to the right of vertex C; vertex E is inclined upward to the left of vertex F; vertex G is inclined upward to the right of vertex E. The edges, represented by line segments between the vertices