By Fazlollah M. Reza

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5 X4 3121 = -"2T- = 10 Binomial Expansion: Let n be a positive integer; then (a + b)n = an + (~) an-1b + (;) an- + . (n - 2) an- + ... + b" 2b 2 3b3 (2-111) A useful display of a binomial coefficient is given in a table which is called Pascal' 8 triangle: 52 DISCRETE SCHEMES 'VITHOUT MEMORY 1 1 121 3 3 1 464 1 In the following a number of simple examples dealing with permutations and combinations are presented. In these examples, the primary assumption is that the probability is given by the frequency of the event under consideration; that is, the concept of equiprobable measure prevails.

The probability of the closing of each relay of the circuit of Fig. E2-19 is a given a. Assuming that all relays act independently, what is the probability of a current existing between terminals A and B? FIG. E2-19 Solution. Let the event of closing each relay 1, 2, 3, and 4 be E I , E 2, E a, and E 4, respectively. The four events are independent but not necessarily mutually exclusive. The event of interest is E = E IE 2 + E aE 4 PtE} =P{E IE 2 + E aE 4 } = P{E IE 2} + P{E aE4 } - P{E IE 2E aE 4 } PtE} = P{EdP{E 2 } + P{E3}P{E~} - P{EdP{E 2}P{E 3}P{E4 } PtE} = 20: 2 - 0: 4 46 DISCRETE SCHEMES WITHOUT MEMORY Note that o :::; PlO} = 0 P{l} = 1 P {E} S 1 for 0 :::; a S 1 In many we wish to concentrate on two mutually exclusive and exhaustive events of the sample space, that is, two events and A 2 such that A lA 2 Al +A 2 = (1 = (2-87) U The assumption is that each of these events has a subevent of special interest to us.

Are normally open relays and A', B', and C' are normally closed relays which are respectively activated by the same controlling source. For instance, when relay is open because of the effect of its activating coil, A' is closed. In order to have a current flow between the terminals M and N, we must have the set of relay operations indicated by ABC + AB'C + A'B'C. With this in mind, the question is to the given network by a less complex equivalent circuit. A (a) (b) FIG. Solution. E2-5 A way of simplifying the above expression is the following: F = C(AB + AB' + A'B') F = C[A(B + B') + A'B'] F = C(AU + A'B') F = C(A + A'B') F = C(A + B') A circuit presentation of this example is illustrated in Fig.