Homework 2: Probability And Set Operations

University:
California State University, Northridge
Course:
MATH 340 | Introduction to Probability and Statistics
Academic year:

2023
Views:

296

Pages:

6
Author:

Nyi Moe

11/19/23, 11:36 PM HW02-Nyi Nyi Moe Htet Student: Nyi Nyi Moe Htet Date: 11/19/23 Instructor: Xiaolong Han Course: CSUN_MATH340_HAN (1) Assignment: HW02 1. Prove the following relation. EF ⊂ E⊂E∪F The statement EF ⊂ E ⊂ E ∪ F is true only when thesub-statements EF ⊂ E and E ⊂ E ∪ F are true. The event EF contains the elements in both E and F. Therefore, the statement"the event EF is contained in eventE" is true. elements in either E or F. Therefore, the statement"the event E is contained in the union of E andF" is true. thesub-statements aretrue, the statement EF ⊂ The event E ∪ F contains the Since both of E ⊂ E ∪ F is true. 2. This question has been removed from this assignment by your instructor; you have received full credit. 3. Prove the following relations. c c F = FE ∪ FE and E ∪ F = E ∪ E F c The intersection FE contains all of the elements that are in F that are also in E. The intersection FE contains all of the elements that c are in F but are not in E. Therefore, FE ∪ FE contains all of the elements that are in F and are either in E or not in E, which is c equivalent to F = FE ∪ FE . The c E ∪ F contains all of the elements that are in F or E. union that are in F but are not in E. The The intersection E F contains all of the elements c union E ∪ E F contains all of the elements that are in F or E. c E ∪ F and E ∪ E F contain all of the elements that are Therefore, c in F or E, which is equivalent to E ∪ F = E ∪ E F. 4. Prove the following relations. ∞ ∪ ∞ Ei F = i=1 ∪ ∞ Ei F and ∩ i=1 ∞ Ei i=1 ∪ F= ∩ Ei ∪ F i=1 ∞ The union ∪ Ei contains all of the elements that are all outcomes that are in Ei for at least one value of i. The intersection i=1 ∞ ∪ Ei F contains all of the elements that are in F that are also in Ei . The intersection Ei F contains all of the elements that are i=1 ∞ in F that are also in Ei . Therefore, the union ∪ Ei F contains all of the elements that are in F that are also in Ei . Therefore, both sides i=1 ∞ in F that are also in Ei , of the equation contain all of the elements that are which is equivalent to ∪ ∞ Ei F = i=1 ∪ Ei F . i=1 ∞ The intersection ∩ Ei contains all of the elements that are all outcomes that are in Ei for every value of i. The union i=1 ∞ ∩ ∞ Ei ∪ F contains all of the elements that are in F or i=1 ∩ The Ei . union Ei ∪ F contains all of the elements that i=1 ∞ are in F or Ei . Therefore, the intersection ∩ i=1 ∞ Ei ∪ F contains all of the elements that are in F or ∩ Ei . i=1 ∞ Therefore, both sides of the equation contain all of the elements that are in F or ∩ Ei , which is equivalent to i=1 ∞ ∩ ∞ Ei i=1 ∪ F= ∩ Ei ∪ F . i=1 https://tdx.acs.pearsonprd.tech/api/v1/print/highered 1/6 11/19/23, 11:36 PM HW02-Nyi Nyi Moe Htet 5. For any sequence of events E1 , E2 , ..., define a new sequence F1 , F2 , ..., of disjoint events(that is, events such that Fi Fj = ∅ whenever n n i ≠ j) such that for all n ≥ 1, ∪Fi = ∪Ei . 1 1 Choose the correct answer below. i=1 A. Fi = ∩ c Ej j=1 i=1 B. Fi = Ei c Ej ∩ j=1 i=1 C. Fi = ∪ c Ej j=1 i=1 D. Fi = Ei ∪ c Ej j=1 https://tdx.acs.pearsonprd.tech/api/v1/print/highered 2/6 11/19/23, 11:36 PM HW02-Nyi Nyi Moe Htet 6. Let B, D, and G be three events in the sample space S. Find expressions sothat, of B, D, and G, the events in parts (a) through (j) occur. (a) Only B occurs. Choose the correct answer below. A. B B. B∪Dc ∪Gc C. BDc Gc D. B∪ Dc Gc (b) Both B and G, but not D, occur. Choose the correct answer below. A. B∪G B. BDc G C. BG D. B∪Dc ∪G (c) At least one of the events occurs. Choose the correct answer below. A. S B. B∪Bc ∪D∪Dc ∪G∪Gc C. BDG D. B∪D∪G (d) At least two of the events occur. Choose the correct answer below. A. BD∪Bc Dc ∪BG∪Bc Gc ∪DG∪Dc Gc B. BDG C. Bc Dc Gc D. BD∪BG∪DG (e) All three events occur. Choose the correct answer below. A. B∪D∪G B. B∪Bc ∪D∪Dc ∪G∪Gc C. S D. BDG (f) None of the three events occurs. Choose the correct answer below. A. S B. Bc Dc Gc C. Bc ∪Dc ∪Gc D. ∅ (g) At most one of the events occurs. Choose the correct answer below. A. B∪Bc ∪D∪Dc ∪G∪Gc B. https://tdx.acs.pearsonprd.tech/api/v1/print/highered 3/6 11/19/23, 11:36 PM c c HW02-Nyi Nyi Moe Htet c c B D G ∪BD G C. B∪D∪G c ∪ c B DG c ∪ c c B D G D. BDG (h) At most two of the events occur. Choose the correct answer below. A. BD∪BG∪DG B. Bc Dc Gc ∪BDc Gc ∪Bc DGc ∪Bc Dc G C. BDc Gc ∪Bc DGc ∪Bc Dc G D. (BDG)c (i) Exactly two of the events occur. Choose the correct answer below. A. Bc Dc Gc ∪BDc Gc ∪Bc DGc ∪Bc Dc G B. BDGc ∪Bc DG∪BDc G C. BD∪BG∪DG D. (BDG)c (j) At most three of the events occur. Choose the correct answer below. A. BDG B. B∪Bc ∪D∪Dc ∪G∪Gc C. S D. B∪D∪G 7. Use the given Venn diagram with events E and F and regions1, 2,3, and 4 to complete parts(a) and(b). E F 3 1 (a) Simplify the expression (E∪F) E∪F ( E∪ F ) E∪ F c = c 2 4 . E c c (b) ProveDeMorgan's laws for events E and F.[That is, prove (E∪F)c = E F and (EF)c = E c ∪ c F .] c c Begin by proving (E∪F)c = E F . Using the Venn diagram in the problemstatement, theregion(s) in (E∪F)c is/are (Type a whole number. Use a comma to separate answers asneeded.) c Theregion(s) in E is/are 2,4 c , and theregion(s) in F is/are 1,4 4 . . The intersection of these two events is theregion(s) in (E∪F)c , thereby completing the proof. equal to (Type whole numbers. Use a comma to separate answers asneeded.) Now prove (EF)c = E c ∪ c F . Theregion(s) in (EF)c is/are 1,2,4 . (Type a whole number. Use a comma to separate answers asneeded.) c c Recall that the regions in E and F were determined previously. The equal to union of these two events is theregion(s) in (EF)c , thereby completing the proof. https://tdx.acs.pearsonprd.tech/api/v1/print/highered 4/6 11/19/23, 11:36 PM HW02-Nyi Nyi Moe Htet 8. Suppose that an experiment is performed n times. For any event E of the samplespace, letn(E) denote the number of times that event E n(E) occurs and define f(E) = . Show that f( • ) satisfies the accompanying three axioms of probability.(Note that the function f( • ) n represents the function f with exactly one undefined variable as itsargument.) 1 Click the icon to view the axioms. Show that f( • ) satisfies Axiom 1. The function f( • ) satisfies Axiom 1 because n( • ) is always greater than nonnegative, and n is always positive and n( • ), which means that dividing n( • ) by n will always result in a value between 0 and1, inclusive. Show that f( • ) satisfies Axiom 2. (S) = n n . As aresult, f(S) = (Simplify youranswers.) 1 . Therefore, f( • ) satisfies Axiom 2. Show that f( • ) satisfies Axiom 3. ∞ ∞ If the events E1 , E2 , ... are mutuallyexclusive, then n ∪ Ei must be equal to i=1 f ∪ n Ei . Dividing both of these expressions by i=1 ∞ ∞ n, ∑ Ei must be equal to i=1 ∑ f Ei . Therefore, f( • ) satisfies Axiom 3. i=1 1: Axioms Consider an experiment whose sample space is S. For each event E of the sample spaceS, assume that a numberP(E) is defined and satisfies the following three listed axioms. Axiom1: 0 ≤ P(E) ≤ 1 Axiom2: P(S) = 1 Axiom3: For any sequence of mutually exclusive events E1 , E2 , ...(that is, events for which Ei Ej ≠ ∅ when i ≠ j), the following equation is true. ∞ ∞ P ∪ Ei = i=1 ∑ P Ei i=1 The numberP(E) is referred to as the probability of the event E. 9. This question has been removed from this assignment by your instructor; you have received full credit. 10. Show that the probability that exactly one of the events E or F occurs equals P(E) + P(F) − 2P(EF). The event"exactly one of the events E or Foccurs" can be written as EF c ∪ c E F. The probability of the event determined in the previous step can be rewritten as P EF The first term can be rewritten as addition P(E) − P(EF) c c +P E F . and the second term can be rewritten as P(F) − P(EF). Performing the between the two rewritten terms and simplifying the expression results in P(E) + P(F) − 2P(EF), thereby completing the proof. https://tdx.acs.pearsonprd.tech/api/v1/print/highered 5/6 11/19/23, 11:36 PM HW02-Nyi Nyi Moe Htet 11. Prove that P EFc = P(E) − P(EF). c The event E can be written as E = EF∪EF . The probability of the event determined in the previous step can be rewritten as P(E) = P(EF) + P EF Taking the value P EF c P(EF), c . subtracting it from both sides of the equation in the previous step, and simplifying results in = P(E) − P(EF), thereby completing the proof. https://tdx.acs.pearsonprd.tech/api/v1/print/highered 6/6