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The Z-parameters of a two-port network N are given by Z11 2s 1=s Z12 Z21 2s Z22 2s 4 (a) Find the T-equivalent of N. (b) The network N is connected to a source and a load as shown in the circuit of Fig. 13-8. Replace N by its T-equivalent and then solve for i1 , i2 , v1 , and v2 .

Method 2 The probability of choosing any red ball is 5>12. Then the probability of choosing 3 red balls is (5 >12)3. Similarly, the probability of choosing 2 white balls is (4 >12)2, and of choosing 1 blue ball, (3 >12)1. Therefore, the probability of choosing 3 red, 2 white, and 1 blue in that order is a 5 3 4 2 3 1 b a b a b 12 12 12

One excellent management practice is error tracking, which is keeping track of the errors that have occurred and the inter-error times (the time between occurrences of the errors). This can be used to make decisions about when to release software. An additional e ect of tracking and publicizing the error rate is to make the

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(a) The three branches of the T-equivalent network (Fig. 13-4) are 1 1 Za Z11 Z12 2s 2s s s Zb Z22 Z12 2s 4 2s 4 Zc Z12 Z21 2s (b) The T-equivalent of N, along with its input and output connections, is shown in phasor domain in Fig. 13-17.

But the same selection can be achieved in various other orders, and the number of these different ways is 6! 3!2!1! as shown in 1. Then the required probability is 6! 5 3 4 2 3 1 a b a b a b 3!2!1! 12 12 12 Method 3 The required probability is the term p3 p2 pb in the multinomial expansion of (pr r w pw 4>12, pb 3>12. By actual expansion, the above result is obtained. pw pb)6 where pr 5>12,

The hypergeometric distribution 4.27. A box contains 6 blue marbles and 4 red marbles. An experiment is performed in which a marble is chosen at random and its color observed, but the marble is not replaced. Find the probability that after 5 trials of the experiment, 3 blue marbles will have been chosen.

Fig. 13-17

Method 1 6 The number of different ways of selecting 3 blue marbles out of 6 blue marbles is a b . The number of different 3 4 ways of selecting the remaining 2 marbles out of the 4 red marbles is a b . Therefore, the number of different 2 6 4 samples containing 3 blue marbles and 2 red marbles is a b a b . 3 2

CHAP. 13]

Now the total number of different ways of selecting 5 marbles out of the 10 marbles (6 10 is a b . Therefore, the required probability is given by 5 6 4 a ba b 3 2 10 21 10 a b 5 Method 2 (using formula) We have b 6, r 4, n 5, x 3. Then by (19), page 113, the required probability is 6 4 a ba b 3 2 10 a b 2

By applying the familiar analysis techniques, including element reduction and current division, to Fig. 13-17, we nd i1 , i2 , v1 , and v2 . In phasor domain I1 I2 V1 V2 3:29 10:28 1:13 131:28 2:88 37:58 1:6 93:88 In the time domain: i1 3:29 cos t 10:28 i2 1:13 cos t 131:28 v1 2:88 cos t 37:58 v2 1:6 cos t 93:88

The uniform distribution 4.28. Show that the mean and variance of the uniform distribution (page 113) are given respectively by 1 1 (a) m 2 (a b), (b) s2 12 (b a)2.

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