HW2Sol

# HW2Sol - TE15 ST14 Since F are mutually exclusive and U F i...

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EE/NIS 605-CS505 Probability and Stochastic Processes I Homework 2 Solutions Professor Heffes Textbook: S. Ross, A First Course in Probability, Seventh Edition , Prentice Hall, 2006 Chapter 2 : TE: 5, 6, 7a, 11,13,15, ST:14, Problems: 25 TE5 : TE6 : (a) EF c G c (b) EF c G (c) E U F U G (d) EF U EG U FG (e) EFG (f) E c F c G c (g) (E c F c G c ) U (EF c G c ) U (E c FG c ) U (E c F c G) (h) (EFG) c (i) (EFG c ) U (EF c G) U (E c FG) (j) S TE7 : (a) (E U F)(E U F c ) = (E U F)E U (E U F) F c = E U(EF c )=E TE11 : P(E U F) = P(E) + P(F) – P(EF) < 1 P(EF) > P(E) + P(F) -1 TE13 : E = EF U EF c , P(E) = P(EF) + P(EF c ) since EF and EF c are disjoint.

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Unformatted text preview: TE15 : ST14 : Since F are mutually exclusive and U F i = U E i , Professor Heffes F i = E i ∪ j = 1 i – 1 E j c = E i ∩ j = 1 i – 1 E j c M k N r – k M + N r F i = E i E i – 1 c ...E 1 c , i > 1, F 1 = E 1 P ∪ i = 1 n E i = P ∪ i = 1 n F i = P(F i ) Σ 1 = 1 n Since F i ⊂ E i , P(F i ) < P(E i ) Σ 1 = 1 n Σ 1 = 1 n P25 : Professor Heffes P[E n ] = 26.26. ...26 .4 36.36. ..36 = ( 26 36 ) n – 1 4 36 Σ n = 1 ∞ P[E n ] = 4 36 Σ n = 1 ∞ ( 26 36 ) n – 1 = 4 36 1 – 26 36 = 4 10 = 2 5...
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HW2Sol - TE15 ST14 Since F are mutually exclusive and U F i...

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