[...]... 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 are independent, then it can be shown that A and B are also independent; that is (Prob 1.53), Then Thus, if A is independent... 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 or s, is closed Thus A,,... 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. .. Verify the distributive law (1.12) Let s E [ A n ( B u C)] Then s E A and s E (B u C) This means either that s E A and s E B or that s E A and s E C; that is, s E ( A n B) or s E ( A n C) Therefore, A n ( B u C ) c [ ( A n B) u ( A n C)] Next, let s E [ ( A n B) u ( A n C)] Then s E A and s E C) Thus, E B or s E A and s E C Thus s E A and (s E B or s [(A n B) u ( A n C)] c A n (B u C) Thus, by the definition... Let B be the event that "the part selected is defective," and let A be the event that "the part selected came from plant 1." Then A n B is the event that the item selected is defective and came from plant 1 Since a part is selected at random, we assume equally likely events, and using Eq (1.38), we have Similarly, since there are 3000 parts and 250 of them are defective, we have By Eq (1.39), the probability... is Eq (1 l9) 16 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... source generating two symbols, dots and dashes We observed that the dots were twice as likely to occur as the dashes Find the probabilities of the dot's occurring and the dash's occurring CHAP 1 1 * PROBABILITY From the observation, we have P(dot) = 2P(dash) Then, by Eq (1.39, P(dot) + P(dash) = 3P(dash) = 1 P(dash) = 5 Thus, 1.30 and P(dot) = The sample space S of a random experiment is given by S =... 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... randomly from a group of 5 men and 10 women (a) Find the probability that the committee consists of 2 men and 3 women (b) Find the probability that the committee consists of all women (a) The number of total outcomes is given by It is assumed that "random selection" means that each of the outcomes is equally likely Let A be the event that the committee consists of 2 men and 3 women Then the number of... will or will not work Find the probability that a closed path will exist between terminals a and b Fig 1-12 CHAP 11 PROBABILITY 23 Consider a sample space S of which a typical outcome is (1,0,0, I), indicating that switches 1 and 4 are closed and switches 2 and 3 are open The sample space contains 24 = 16 points, and by assumption, they are equally likely (Fig 1-13) Let A,, i = 1, 2, 3, 4 be the event . 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. of Random Variables, Expectation, Limit Theorems 122 4.1 Introduction 122 4.2 Functions of One Random Variable 122 4.3 Functions of Two Random Variables 123 4.4 Functions of n Random Variables. 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 by The McGraw-Hill