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# 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 ) = 1 / 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 ) = 13 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 fi rst card can be anything. Given a particular suit on the fi rst card, the probability of suit match on the second card is 12 / 51 ; on the third card it is 11 / 50 ; on the fourth card it is 10 / 49 ; on the fi 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 fl 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 = 0 . 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 ) . The only way that both of these conditions can be true is for P ( A ) = P ( B ) = 0 . Problem 4.5 (a) The result is P ( AB ) = 1 P (1 or more links broken) = 1 ¡ 1 q 2 ¢ 2 (1 q ) (b) If link 4 is removed, the result is P ( AB | link 4 removed ) = 1 ¡ 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 = 0 . 1 Problem 4.7 (a) P ( A 2 ) = 0 . 3; P ( A 2 , B 1 ) = 0 . 05; P ( A 1 , B 2 ) = 0 . 05; P ( A 3 , B 3 ) = 0 . 05; P ( B 1 ) = 0 . 15; P ( B 2 ) = 0 . 25; P ( B 3 ) = 0 . 6 . (b) P ( A 3 | B 3 ) = 0 . 083; P ( B 2 | A 1 ) = 0 . 091; P ( B 3 | A 2 ) = 0 . 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 , jumps by 1/8 at x = 0 , jumps another 3/8 at x = 1 , jumps another 3/8 at x = 2 , and jumps another 1/8 at x = 3 . The pdf consists of an impulse of weight 1/8 at x = 0 , impulses of weights 3/8 at x = 1 and 2, and an impulse of weight 1/8 at x = 3 .
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