Hw2sol_F10 - available] = 1- P[ F 1 F 2 F 3 ]= 1-0.005 =...

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Online EE 131A Homework #2 Fall 2010 Solution K. Yao 1. There are 10 choices for plant A, but only 9 for plant B, and 8 for plant C. This give a total of P 10 3 = 10! / 7! = 10 × 9 × 8 = 720 ways of assigning employees to the plants. 2. The number of ordering is the permutation of four things, taking 4 at the first step, 3 at the second step,,etc. Thus, P 4 4 = 4! / 0! = 4! = 4 × 3 × 2 × 1 = 24. 3. Here order is not important and we only want to know how many subsets of size 3 can be selected from 10 people. Thus, C 10 3 = 10! / (3!7!) = (10 × 9 × 8) / (1 × 2 × 3) = 120. 4. a. F E . b. E . c. . d. { O ∪ { 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 , 28 , 30 }} . 5. a. All kings and all clubs. b. King of clubs. c. All kings and all the spades, hearts, and diamonds. d. All spades, hearts, and diamonds except their three kings. e. All kings. 6. Denote F i , i = 1 , 2 , 3 , when frequency F i is available, and ¯ F i when it is not available. Then the sample space S = { 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 , ¯ F 1 ¯ F 2 ¯ F 3 } . a. We are given P[No frequency is available] = P[ ¯ F 1 ¯ F 2 ¯ F 3 ]= 0.005. Thus, P[OK in emergency mode]= P[One or more frequencies available]= 1- P[No frequency is
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Unformatted text preview: available] = 1- P[ F 1 F 2 F 3 ]= 1-0.005 = 0.995. b. We are given P[Losing exactly two frequencies]=P[Only one frequency is avail-able]= 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 fre-quency]= 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 ]= 1-P[ { 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 } = 1-0.335 = 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.

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