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. Shaun Sanders 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(XOF) P(X|F) = P(F P(X NF) = P(XNF)P(F) P(XNF)=P(X|F)P(F) Combine and re-arrange: 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(XOF) P(F|X) = P(F) P(XnF)=P(X|F)P(F) P(Fnx) P(X|F) = P(X) Since P(FnX) = P(X n F) because P(X|F) = P(X) PXNF) P(FnX) = P P(X|F)P(F) P(X) 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 = outside, I = inside, L = log everyday P(O)=0.7, P(L|I) = 0.8 = P(L|Õ) P(I) = P(Õ) = 1 -0.7 = 0.3 P(L)= P(L|O)P(O)+P(L|Õ)P(Õ)=(0.6*0.7)+(0.8*0.3)= 0.66 P(L) is 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) = A Probability they are inside the country is 4/11. 0.66
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 are as follows: 3 vertical edges connect the following vertices: B and C, H and I, and H and J; 5 inclined edges connect the following vertices: A and B, C and D, D and E, E and F, and E and G. Determine the properties of the Hasse diagram based on the following questions: (a)