[...]... P(B) - P(A n B) =: 0 9 + 0 8 - 0.75 = 0.95 (b) By Eq (1.64)(Prob 1.23), we have P ( A n B) = P(A) - P(A n B) = 0.9 - 0.75= 0.15 (c) By De Morgan's law, Eq (1.14) ,and Eq (1.25) and using the result from part (a),we get P(A n B) = P(A u B) = 1 - P(A u B) = 1 - 0.95 = 0.05 B) ;and (c) PROBABILITY [CHAP 1 1.25 For any three events A,, A , , and A , , show that + P(A,) + P(A,) - P(A, n A,) P(Al n A,) - P(A,... n n A ,- ,) Multiplying both sides by P(A,+, I A , n A , n n A,), we have P(Al n A, n and - - - n A,)P(A,+,IA, n A, n A,) = P(Al n A , n P(A, n A , n - n A,, , ) = P(A,)P(A, 1 A,)P(A31 A , rl A,) Thus, Eq (1.81) is also true for n for n 2 2 1.45 n =k - n - - P(A,+, 1 A , A,,,) n A, n - n A,) + 1 By Eq ( 1 A l ) , Eq (1.81) is true for n = 2 Thus Eq (1.81) is true Two cards are drawn at random from... PROBABILITY [CHAP 1 THE NOTION AND AXIOMS OF PROBABILITY 1.18 Using the axioms of probability, prove Eq (1.25) We have S = A u A and AnA=@ Thus, by axioms 2 and 3, it follows that P(S) = 1 = P(A) + P(A) from which we obtain P(A) = 1 - P(A) 1.19 Verify Eq (1.26) From Eq (1Z), have we P(A) = 1 - P(A) Let A = @ Then, by Eq (1.2),A = @ = S, and by axiom 2 we obtain P(@)=l-P(S)=l-1=0 1.20 Verify Eq (1.27) Let... s,, and s, are closed, respectively Let A,, denote the event that there is a closed path between terminals a and b Express A,, in terms of A,, A,, and A, for each of the networks shown (4 (b) Fig 1-5 From Fig 1-5 (a), we see that there is a closed path between a and b only if all switches s,, s,, and s, are closed Thus, A,, = A, n A, (3 A, From Fig 1-5 (b), we see that there is a closed path between a and. .. v A, From Fig 1-5 (c), we see that there is a closed path between a and b if s, and either s, or s, are closed Thus, A,, = A, n (A, v A,) Using the distributive law (1.12), we have A,, = (A1 n A,) u (A, n A,) which indicates that there is a closed path between a and b if s, and s, or s, and s, are closed From Fig 1-5 (d), we see that there is a closed path between a and b if either s, and s, are closed... birthday Then B = 2 and by Eq (1.25), P(B) = 1 - P(A) (b) Substituting n = 50 in Eq (1.78), we have P(A) z 0.03 and P(B) z 1 - 0.03 = 0.97 (c) From Eq (1.78), when n = 23, we have P(A) x 0.493 and P(B) = 1 - P(A) w 0.507 That is, if there are 23 persons in a room, the probability that at least two of them have the same birthday exceeds 0.5 1.33 A committee of 5 persons is to be selected randomly from a... right-hand side are all conditioned on events Ai, while the term on the left is conditioned on B Equation (1.45) is sometimes referred to as Bayes' theorem 1.8 INDEPENDENT EVENTS Two events A and B are said to be (statistically) independent if and only if It follows immediately that if A and B are independent, then by Eqs (1.39) and (1.40), P(A I B) = P(A) and P(B I A) = P(B) (1.47) If two events A and. .. B) = P(A) and From Eq (l.61),we have P(An B) = P(B) - P(A n B) Substituting Eq (1.62)into Eq (1.60),we obtain P(A u B) = P(A) + P(B) - P(A n B) CHAP 11 PROBABILITY Shaded region: A n B Shaded region:A n B Fig 1-8 1.22 Let P(A) = 0.9 and P(B) = 0.8 Show that P(A n B) 2 0.7 From Eq (l.29), we have P(A n B) = P(A) + P(B) - P(A u B) By Eq (l.32), 0 I P(A u B) I1 Hence P(A r\ B) 2 P(A) + P(B) - 1 Substituting... events A u B, A n B, and A The Venn diagram in Fig 1-2 indicates that B c A and the event A n B is shown as the shaded area PROBABILITY ( t r ) Shaded ( h )Shaded region: A n B region: A u H (I.) Shaded region: A Fig 1-1 R c A Shaded region: A n R Fig 1-2 C Identities: By the above set definitions or reference to Fig 1-1 , we obtain the following identities: S=@ B=s J=A The union and intersection operations... a Venn diagram, repeat Prob 1.12 Figure 1-6 shows the sequence of relevant Venn diagrams Comparing Fig 1-6 (b) and 1-6 (e), we conclude that ( u ) Shaded region: H ( c ) Shaded u C' ( h )Shaded region: A n ( B u C ) ((1) Shaded region: A n C region: A n H ( r ) Shaded region: (A n H ) u( A n C ) Fig 1-6 1.14 Let A and B be arbitrary events Show that A c B if and only if A n B = A "If" part: We show that . and Digital Communications and Schaum's Outline of Signals and Systems. Schaum's Outline of Theory and Problems of PROBABILITY, RANDOM VARIABLES, AND RANDOM PROCESSES Copyright © 1997. 2. Random Variables 38 2.1 Introduction 38 2.2 Random Variables 38 2.3 Distribution Functions 39 2.4 Discrete Random Variables and Probability Mass Functions 41 2.5 Continuous Random Variables. outline of theory and problems of probability, random variables, and random processes / Hwei P. Hsu. p. cm. — (Schaum's outline series) Includes index. ISBN 0-0 7-0 3064 4-3 1. Probabilities—Problems,