应用随机过程11版（答案全）.pdfVIP

Instructor’s Manual to Accompany Chapter 1 1. S = {(R , R ), (R , G), (R , B ), (G, R ), (G, G), (G, B ), (B , R ), (B , G), (B , B )} The probability of each point in S is 1/9. 2. S = {(R , G), (R , B ), (G, R ), (G, B ), (B , R ), (B , G)} 3. S = {(e , e , . . . , e ), n ≥ 2} where e ∈(heads, tails}. In addition, e = e = 1 2 n i n n −1 heads and for i = 1, . . . , n − 2 if ei = heads, then ei +1 = tails. P {4 tosses} = P {(t , t , h , h)} + P {(h , t , h , h)} = 2 14 = 1 2 8 4. (a) F (E ∪ G)c = F E c Gc (b) E FG c (c) E ∪ F ∪ G (d) E F ∪ EG ∪ FG (e) E FG (f) (E ∪ F ∪ G)c = E c F c Gc c c c (g) (E F ) (EG ) (FG ) (h) (E FG )c 5. 3 . If he wins, he only wins $1, while if he loses, he loses $3. 4 6. If E (F ∪ G) occurs, then E occurs and either F or G occur; therefore, either E F or EG occurs and so E (F ∪ G) ⊂ E F ∪ EG Introduction to Probability Models, Eleventh Edition. /10.1016/B978-0-12-407948-9.00019-0 © 2014 Elsevier Inc. All rights reserved. 2 Introduction to Probability Models Similarly, if E F ∪ EG occurs, then either E F or EG occurs. Thus, E occurs and either F or G occurs; and so E (F ∪ G) occurs. Hence, E F ∪ EG ⊂ E (F ∪ G) which together with the reverse inequality proves the result. 7. If (E ∪ F )c occurs, then E ∪ F does not occur, and so E does not occur (and so E c does); F does not occur (and so F c does) and thus E c and F c both occur. Hence, (E ∪ F )c ⊂ E c F c If E c F c occurs, then E c occurs (and so E does not), and F c occurs (and so F does not). Hence,