chap-4 - Chapter 4 Problems 4 2 6 P{X = 4 = = 14 91 2 4 2 2...

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46 Chapter 4 Chapter 4 Problems 1. P { X = 4} = 91 6 2 14 2 4 = P { X = 0} = 91 1 2 14 2 2 = P { X = 2} = 91 8 2 14 1 2 2 4 = P { X = 1} = 91 16 2 14 1 2 1 8 = P { X = 1} = 91 32 2 14 1 8 1 4 = P { X = 2} = 91 28 2 14 2 8 = 2. p (1) = 1/36 p (5) = 2/36 p (9) = 1/36 p (15) = 2/36 p (24) = 2/36 p (2) = 2/36 p (6) = 4/36 p (10) = 2/36 p (16) = 1/36 p (25) = 1/36 p (3) = 2/36 p (7) = 0 p (11) = 0 p (18) = 2/36 p (30) = 2/36 p (4) = 3/36 p (8) = 2/36 p (12) = 4/36 p (20) = 2/36 p (36) = 1/36 4. P { X = 1} = 1/2, P { X = 2} = 18 5 9 5 10 5 = , P { X = 3} = 36 5 8 5 9 4 10 5 = , P { X = 4} = 168 10 7 5 8 3 9 4 10 5 = , P { X = 5} = 252 5 6 5 7 8 9 10 2 3 4 5 = , P { X = 6} = 252 1 6 7 8 9 10 1 2 3 4 5 = 5. n 2 i , i = 0, 1, …, n 6. P ( X = 3} = 1/8, P { X = 1} = 3/8, P { X = 1} = 3/8, P { X = 3} = 1/8 8. (a) p (6) = 1 (5/6) 2 = 11/36, p (5) = 2 1/6 4/6 + (1/6) 2 = 9/36 p (4) = 2 1/6 3/6 + (1/6) 2 = 7/36, p (3) = 2 1/6 2/6 + (1/6) 2 = 5/36 p (2) = 2 1/6 1/6 + (1/6) 2 = 3/36, p (1) = 1/36 (d) p (5) = 1/36, p (4) = 2/36, p (3) = 3/36, p (2) = 4/36, p (1) = 5/36 p (0) = 6/36, p ( j ) = p ( j ), j > 0
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Chapter 4 47 11. (a) P {divisible by 3} = 1000 333 P {divisible by 105} = 1000 9 P {divisible by 7} = 1000 142 P {divisible by 15} = 1000 66 In limiting cases, probabilities converge to 1/3, 1/7, 1/15, 1/10 (b) P { µ ( N ) 0} = P { N is not divisible by 2 i p , i 1} = i N P { is not divisible by 2 i p } = i i p 2 / 1 1 () = 6 / π 2 13. p (0) = P {no sale on first and no sale on second} = (.7)(.4) = .28 p (500) = P {1 sale and it is for standard} = P {1 sale}/2 =[ P {sale, no sale} + P {no sale, sale}]/2 = [(.3)(.4) + (.7)(.6)]/2 = .27 p (1000) = P {2 standard sales} + P {1 sale for deluxe} = (.3)(.6)(1/4) + P {1 sale}/2 = .045 + .27 = .315 p (1500) = P {2 sales, one deluxe and one standard} = (.3)(.6)(1/2) = .09 p (2000) = P {2 sales, both deluxe} = (.3)(.6)(1/4) = .045 14. P { X = 0} = P {1 loses to 2} = 1/2 P { X = 1} = P {of 1, 2, 3: 3 has largest, then 1, then 2} = (1/3)(1/2) = 1/6 P { X = 2} = P {of 1, 2, 3, 4: 4 has largest and 1 has next largest} = (1/4)(1/3) = 1/12 P { X = 3} = P {of 1, 2, 3, 4, 5: 5 has largest then 1} = (1/5)(1/4) = 1/20 P { X = 4} = P {1 has largest} = 1/5
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48 Chapter 4 15. P { X = 1} = 11/66 P { X = 2} = + = j j j 54 11 66 12 11 2 P { X = 3} = + + + ∑∑ ≠= k j j k j j k kj 42 11 54 12 66 12 12 P { X = 4} = 1 = = 3 1 } 1 { i X P 16. P { Y 1 = i } = 66 12 i P { Y 2 = i } = + i j j i j 54 12 66 12 P { Y 3 = i } = + + + ≠≠ j k j k j i k j ki j 42 11 54 12 66 12 All sums go from 1 to 11, except for prohibited values. 20. (a) P { x > 0} = P {win first bet} + P {lose, win, win} = 18/38 + (20/38)(18/38) 2 .5918 (b) No, because if the gambler wins then he or she wins $1. However, a loss would either be $1 or $3.
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chap-4 - Chapter 4 Problems 4 2 6 P{X = 4 = = 14 91 2 4 2 2...

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