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.
Jaime Rowland
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 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.
Proof: P(x)= P(x-F).P(F) /= P(X-F).P(F) Because of conditional probability we can say -P(X-F)=P(X nF)/P(F)= P(Xn F)= P(X|F).P(F)(1)= P(X n F) = =(x)d=()d(|x)d=(UN)d+(Ux)d=()+(1)iMON(z)+()d(|x) P(X|F).P(F)+P(X|F).P(F)Thereforethisisprovedbecausethesolutionisthesameasthefigure
ExplainwhyP(F-X)=P(X-F)P(F)/P(X)isanotherwayof statingTheorem4.2.1Bayes'Theorem.
P(X|F)=P(XnF)/P(F)=P(XnF)=P(X|F).P(F)(1)
P(F-X)=P(NnF)/P(X)=P(F|X)=P(X|F)P(F)/P(X)USING(1)=P(F|X)=P(X|F).P(F)/P(X|F)P(F)+ P(X|F).P(F)(USINGA)THISISTHEBAYESTHEOREM.
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.
(b) What percent of all users log on every day? Hint: Use the equation from Part 1 (a)
P(L) = (PO).P(O) + P(LO0.P(O) =(0.60 X 0.70) + (0.80 X 0.30) = 0.42 + 0.24 =P(L) = 0.66 = 66I.E. 66 PERCENT OF ALL USERS LOG ON EVERY DAY
(b) Using Bayes' Theorem, out of users who log on every day, what is the probability that they are from inside the country?
P(IL) = P(I).P(LI)/P(0).P(L0) + P(I).P(LI) =0.30 X 0.80/0.66 =P(I/L) = 4/11 THIS IS THE REQUIRED PROBABILITY
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
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) What are the minimal elements of the partial order?
AES IS MINIMAL IN (S,<IFTHEREISNOBESSUCHTHATA< B,FROMTHETOPOFTHEHASSLE I,A,F)
Whatarethemaximalelementsofthepartialorder?
AESISMINIMALIN(S,)IFBESSUCHTHATBAESUCHTHATB AFROMTHEBOTTOMOFTHE
(b) Which of the following pairs are comparable? (A, D), (J, F), (B, E), (G, F), (D,B), (C, F), (H,I),(C,E)
(A,D), (G,