MIT6_041F10_assn01_sol

# MIT6_041F10_assn01_sol - 0000000 0000000 0000000 0000000...

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Unformatted text preview: 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) Problem Set 1: Solutions Due: September 15, 2010 1. (a) A ∪ B ∪ C (b) ( A ∩ B c ∩ C c ) ∪ ( A c ∩ B ∩ C c ) ∪ ( A c ∩ B c ∩ C ) ∪ ( A c ∩ B c ∩ C c ) (c) ( A ∪ B ∪ C ) c = A c ∩ B c ∩ C c (d) A ∩ B ∩ C (e) ( A ∩ B c ∩ C c ) ∪ ( A c ∩ B ∩ C c ) ∪ ( A c ∩ B c ∩ C ) (f) A ∩ B ∩ C c (g) A ∪ ( A c ∩ B c ) Ω 1111111 1111111 1111111 1111111 1111111 1111111 1111111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 A B C A B C Ω A B C Ω A C Ω B (a) (b) (c) (d) A C Ω B A C Ω B B A Ω (e) (f) (g) 2. Since all outcomes are equally likely we apply the discrete uniform probability law to solve the problem. To solve for any event we simply count the number of elements in the event and divide by the total number of elements in the sample space. There are 2 possible outcomes for each ﬂip, and 3 ﬂips. Thus there are 2 3 = 8 elements (or sequences) in the sample space. (a) Any sequence has probability of 1/8. Therefore P ( { H,H,H } ) = 1/8 . (b) This is still a single sequence, thus P ( { H,T,H } ) = 1/8 . (c) The event of interest has 3 unique sequences, thus P ( { HHT,HTH,THH } ) = 3/8 . (d) The sequences where there are more heads than tails are A : { HHH,HHT,HTH,THH } . 4 unique sequences gives us P ( A ) = 1/2 . 3. The easiest way to solve this problem is to make a table of some sort, similar to the one below. Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) Die 1 1 1 1 1 2 2 2 2 3 3 3 3...
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## This note was uploaded on 01/11/2012 for the course EE 6.431 taught by Professor Prof.dimitribertsekas during the Fall '10 term at MIT.

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MIT6_041F10_assn01_sol - 0000000 0000000 0000000 0000000...

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