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Unformatted text preview: Probability and Stochastic Processes Homework 1 Solutions 29th Jan, Spring’10 Problem Solutions : Yates and Goodman, 1.1.1 1.2.1 1.2.3 1.3.1 1.3.2 1.3.5 1.4.1 1.4.5 1.4.6 1.5.1 1.5.3 1.5.4 1.5.5 1.5.6 and 1.6.1 Problem 1.1.1 Solution Based on the Venn diagram M O T the answers are fairly straightforward: (a) Since T ∩ M negationslash = 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 1.2.1 Solution (a) An outcome specifies whether the fax is high ( h ), medium ( m ), or low ( 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 = { mt,mf } . (c) The event that a fax has two pages is A 2 = { ht,mt,lt } . (d) The event that a fax is either high speed or low speed is A 3 = { ht,hf,lt,lf } . 1 (e) Since A 1 ∩ A 2 = { mt } and is not empty, A 1 , A 2 , and A 3 are not mutually exclusive. (f) Since A 1 ∪ A 2 ∪ A 3 = { ht,hf,mt,mf,lt,lf } = S, (2) the collection A 1 , A 2 , A 3 is collectively exhaustive. 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 1.3.1 Solution The sample space of the experiment is S = { LF,BF,LW,BW } . (1) From the problem statement, we know that P [ LF ] = 0 . 5, P [ BF ] = 0 . 2 and P [ BW ] = 0 . 2. This implies P [ LW ] = 1 − . 5 − . 2 − . 2 = 0 . 1. The questions can be answered using Theorem 1.5. (a) The probability that a program is slow is P [ W ] = P [ LW ] + P [ BW ] = 0 . 1 + 0 . 2 = 0 . 3 . (2) (b) The probability that a program is big is P [ B ] = P [ BF ] + P [ BW ] = 0 . 2 + 0 . 2 = 0 . 4 . (3) (c) The probability that a program is slow or big is P [ W ∪ B ] = P [ W ] + P [ B ] − P [ BW ] = 0 . 3 + 0 . 4 − . 2 = 0 . 5 . (4) Problem 1.3.2 Solution A sample outcome indicates whether the cell phone is handheld ( H ) or mobile ( M ) and whether the speed is fast (...
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 Spring '08
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 Conditional Probability, Probability, Probability theory

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