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Problem Solutions – Chapter 1 Problem 1.1.1 Solution Based on the Venn diagram M O T the answers are fairly straightforward: (a) Since T M 6= φ , T and M are not mutually exclusive. (b) Every pizza is either Regular ( R ), or Tuscan ( T ). Hence R T = S so that R and T are collectively exhaustive. Thus its also (trivially) true that R T M = S . That is, R , T and M are also collectively exhaustive. (c) From the Venn diagram, T and O are mutually exclusive. In words, this means that Tuscan pizzas never have onions or pizzas with onions are never Tuscan. As an aside, “Tuscan” is a fake pizza designation; one shouldn’t conclude that people from Tuscany actually dislike onions. (d) From the Venn diagram, M T and O are mutually exclusive. Thus Gerlanda’s doesn’t make Tuscan pizza with mushrooms and onions. (e) Yes. In terms of the Venn diagram, these pizzas are in the set ( T M O ) c . Problem 1.1.2 Solution Based on the Venn diagram, M O T the complete Gerlandas pizza menu is Regular without toppings Regular with mushrooms Regular with onions Regular with mushrooms and onions Tuscan without toppings Tuscan with mushrooms 2
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Problem 1.2.1 Solution (a) An outcome specifies whether the fax is high ( h ) , medium ( m ) ,orlow ( l ) speed, and whether the fax has two ( t ) pages or four ( f ) pages. The sample space is S = { ht , hf , mt , mf , lt , lf } . (1) (b) The event that the fax is medium speed is A 1 ={ , } . (c) The event that a fax has two pages is A 2 , , } . (d) The event that a fax is either high speed or low speed is A 3 , , , } . (e) Since A 1 A 2 } and is not empty, A 1 , A 2 , and A 3 are not mutually exclusive. (f) Since A 1 A 2 A 3 = { , , , , , } = S , (2) the collection A 1 , A 2 , A 3 is collectively exhaustive. Problem 1.2.2 Solution (a) The sample space of the experiment is S = { aaa , aaf , afa , faa , ffa , faf , aff , fff } . (1) (b) The event that the circuit from Z fails is Z F = { , , , } . (2) The event that the circuit from X is acceptable is X A = { , , , } . (3) (c) Since Z F X A , } 6= φ , Z F and X A are not mutually exclusive. (d) Since Z F X A , , , , , S , Z F and X A are not collectively exhaustive. (e) The event that more than one circuit is acceptable is C = { , , , } . (4) The event that at least two circuits fail is D = { , , , } . (5) (f) Inspection shows that C D = φ so C and D are mutually exclusive. (g) Since C D = S , C and D are collectively exhaustive. 3
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Problem 1.2.3 Solution The sample space is S = { A ,..., K , A K , A K , A K ♠} . (1) The event H is the set H = { A K ♥} . (2) Problem 1.2.4 Solution The sample space is S = 1 / 1 ... 1 / 31 , 2 / 1 2 / 29 , 3 / 1 3 / 31 , 4 / 1 4 / 30 , 5 / 1 5 / 31 , 6 / 1 6 / 30 , 7 / 1 7 / 31 , 8 / 1 8 / 31 , 9 / 1 9 / 31 , 10 / 1 10 / 31 , 11 / 1 11 / 30 , 12 / 1 12 / 31 . (1) The event H defined by the event of a July birthday is described by following 31 sample points. H = { 7 / 1 , 7 / 2 7 / 31 } . (2) Problem 1.2.5 Solution Of course, there are many answers to this problem. Here are four event spaces.
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This note was uploaded on 12/12/2011 for the course EE EE980 taught by Professor Ji during the Spring '11 term at Akademia Rolnicza w Wrocławiu.

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