chapter 1

Chapter 2: Probability79.Using the hints, let P(A i ) = p, and x = p 2 , then P(system lifetime exceeds t 0 ) = p 2 + p 2 – p 4 =2p 2 – p 4 = 2x – x 2 . Now, set this equal to .99, or 2x – x 2 = .99 ⇒ x 2 – 2x + .99 = 0. Use the2 ±4 − (4)(.99)2 ± .22quadratic formula to solve for x: = = = 1±. 1 = .99 or 1.01Since the value we want is a probability, and has to be = 1, we use the value of .99.280. Event A: { (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) }, P(A) = 6 1 ;Event B: { (1,4)(2,4)(3,4)(4,4)(5,4)(6,4) }, P(B) = 6 1 ;Event C: { (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) }, P(C) = 6 1 ;Event A∩B: { (3,4) }; P(A∩B) = 1 36;Event A∩C: { (3,4) }; P(A∩C) = 1 36;Event B∩C: { (3,4) }; P(A∩C) = 1 36;Event A∩B∩C: { (3,4) }; P(A∩B∩C) = 1 36;1 1⋅ =P(A)⋅P(B)= =P(A∩B)6 6 361 1⋅ =P(A)⋅P(C)= =P(A∩C)6 6 361 1⋅ =P(B)⋅P(C)= =P(B∩C)6 6 36The events are pairwise independent.1111 1 1 1⋅ ⋅ = ≠P(A)⋅P(B) ⋅P(C)= = P(A∩B∩C)6 6 6 216 36The events are not mutually independent183