ch04 - Chapter 4 Probability and Random Variables 4.1...

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Chapter 4 Probability and Random Variables 4.1 Problem Solutions Problem 4.1 S = sample space is the collection of the 25 parts. Let A i = event that the pointer stops on the i th part, i =1 , 2 ,..., 25 . These events are exhaustive and mutually exclusive. Thus P ( A i )=1 / 25 (a) P (even number) = P (2 or 4 or ··· or 24) = 12 / 25 ; (b) P ( A 25 / 25 ; (c) P (4 or 5 or 9) = 3 / 25 ; (d) P (number > 10 ) = P (11 , 12 , , or 25) = 15 / 25 = 3 / 5 . Problem 4.2 (a) Use a tree diagram similar to Fig. 4.2 to show that P (3 K, 2 A )= 4 52 ½ 4 51 · 3 3 50 3 49 2 48 ¸ + 3 51 · 3 4 50 3 49 2 48 ¸ + 3 51 · 4 50 3 49 2 48 ¸ + 4 51 · 3 3 50 3 49 2 48 ¸¾ =9 . 2345 × 10 6 (b) Use the tree diagram of Fig. 4.2 except that we now look at outcomes that give 4 of a kind. Since the tree diagram is for a particular denomination, we multiply by 13 and get P (4 of a kind 1 3 ½ 4 52 · 3 51 μ 3 2 50 1 49 48 48 + 48 51 3 50 2 49 1 48 ¸ + 48 52 4 51 3 50 2 49 1 48 ¾ =0 . 0002401 1

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2 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES (c) The f rst card can be anything. Given a particular suit on the f rst card, the probability o fsu itmatchonthesecondcardis 12 / 51 ;ontheth irdcarditis 11 / 50 ;onthefourthcard it is 10 / 49 ;onthe f fth and last card it is 9 / 48 . Therefore, P ( all same suit )=1 12 51 11 50 10 49 9 48 =0 . 001981 (d) For a royal F ush P ( A,K,Q,J, 10 of same suit )= 4 52 1 51 1 50 1 49 1 48 =1 . 283 × 10 8 (e) The desired probability is P ( Q | A,K,J, 10 not all of same suit 4 48 . 833 Problem 4.3 P ( A,B P ( A ) P ( B ) P ( A,C P ( A ) P ( C ) P ( B,C P ( B ) P ( C ) P ( A,B,C P ( A ) P ( B ) P ( C ) Problem 4.4 A and B being mutually exclusive implies that P ( A | B P ( B | A )=0 . A and B being statistically independent implies that P ( A | B P ( A ) and P ( B | A P ( B ) .T h eo n l y way that both of these conditions can be true is for P ( A P ( B . Problem 4.5 (a) The result is P ( AB P (1 or more links broken) ¡ 1 q 2 ¢ 2 (1 q ) (b) If link 4 is removed, the result is P ( AB | link 4 removed ¡ 1 q 2 ¢ (1 q )
4.1. PROBLEM SOLUTIONS 3 (c) If link 2 is removed, the result is P ( AB | link 2 removed )=1 ¡ 1 q 2 ¢ 2 (d) Removal of link 4 is more severe than removal of link 2. Problem 4.6 Using Bayes’ rule P ( A | B )= P ( B | A ) P ( A ) P ( B ) where, by total probability P ( B P ( B | A ) P ( A )+ P ¡ B | A ¢ P ¡ A ¢ = P ( B | A ) P ( A £ 1 P ¡ B | A ¢¤ P ¡ A ¢ =( 0 . 9) (0 . 4) + (0 . 4) (0 . 6) = 0 . 6 Therefore P ( A | B (0 . 9) (0 . 4) 0 . 6 =0 . 6 Similarly, P ¡ A | B ¢ = P ¡ B | A ¢ P ( A ) P ¡ B ¢ with P ¡ B ¢ = P ¡ B | A ¢ P ( A P ¡ B | A ¢ P ¡ A ¢ =(0 . 1) (0 . 4) + (0 . 6) (0 . 6) = 0 . 4 Thus P ¡ A | B ¢ = (0 . 1) (0 . 4) 0 . 4 . 1 Problem 4.7 (a) P ( A 2 )=0 . 3; P ( A 2 ,B 1 . 05; P ( A 1 2 . 05; P ( A 3 3 . 05; P ( B 1 . 15; P ( B 2 0 . 25; P ( B 3 . 6 . (b) P ( A 3 | B 3 . 083; P ( B 2 | A 1 . 091; P ( B 3 | A 2 . 333 . Problem 4.8 See the tables below for the results.

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4 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES (a) P spots up X 1 P ( X 1 = x i ) 2 4 1/36 3 9 2/36 4 16 3/36 5 25 4/36 6 36 5/36 7 49 6/36 8 64 5/36 9 81 4/36 10 100 3/36 11 121 2/36 12 144 1/36 (b) P spots up X 2 P ( X 2 = x i ) 2 1 3 1 4 1 15/36 5 1 6 1 7 0 8 0 9 0 10 0 21/36 11 0 12 0 Problem 4.9 The cdf is zero for x< 0 ,jump sby1/8a t x =0 , jumps another 3/8 at x =1 s another 3/8 at x =2 ,andjumpsanother1/8at x =3 . The pdf consists of an impulse of weight 1/8 at x ,impu lseso fwe ights3/8at x and 2, and an impulse of weight 1/8 at x .
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This note was uploaded on 04/28/2008 for the course EE ELCT 564 taught by Professor Ali during the Spring '08 term at University of South Carolina Beaufort.

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ch04 - Chapter 4 Probability and Random Variables 4.1...

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