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Unformatted text preview: available] = 1 P[ F 1 F 2 F 3 ]= 10.005 = 0.995. b. We are given P[Losing exactly two frequencies]=P[Only one frequency is available]= P[ F 1 F 2 F 3 ]= P[ F 1 F 2 F 3 ]= P[ F 1 F 2 F 3 ] = 0.01 and P[Losing exactly one frequency]= P[Exactly two frequencies are available] = P[ F 1 F 2 F 3 ] = P[ F 1 F 2 F 3 ] = P[ F 1 F 2 F 3 ]= 0.1. This means P[Three frequencies are available] = P[ F 1 F 2 F 3 ]= 1P[ { F 1 F 2 F 3 , F 1 F 2 F 3 , F 1 F 2 F 3 , F 1 F 2 F 3 , F 1 F 2 F 3 , F 1 F 2 F 3 , F 1 F 2 F 3 } ] = 1 3 . 1 3 . 01 . 005 } = 10.335 = 0.665. Thus, P[OK in normal mode]= P[Two or more frequencies are available] = P[Two frequencies are available] + P[Three frequencies are available] = 3 . 1 + 0 . 665 = 0.965....
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This note was uploaded on 11/05/2010 for the course ELECTRICAL EE131A taught by Professor Kungyao during the Fall '10 term at UCLA.
 Fall '10
 KungYao

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