# chapter18 - 1(a = 2 arrivals/hr = 3 services/hr = 2 3 P...

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Unformatted text preview: 1 Chapter 18 181. (a) = 2 arrivals/hr, = 3 services/hr, = 2 / 3. P ( n &amp;gt; 5) = 1- P ( n 5) = 1- 5 X j =0 1 3 2 3 j = 0 . 088 (b) L = 1- = 2, L q = 2 ( - ) = 4 / 3 (c) W = 1 - = 1 hour 182. (a) State 0 = rain; State 1 = clear. P = . 7 0 . 3 . 1 0 . 9 (b) p = 0 . 7 p + 0 . 1 p 1 1 = p + p 1 This implies p = 1 / 4, p 1 = 3 / 4. (c) To find p (3) 11 , note that P (3) = P 3 = . 412 0 . 588 . 196 0 . 804 Thus, p (3) 11 = 0 . 804. (d) f (2) 10 = p (2) 10- f (1) 10 p 10 = 0 . 16- (0 . 1)(0 . 1) = 0 . 15 (e) = 1 /p = 4 days 183. P = p 1- p 1- p p P = 1/2 1/2 1/2 1/2 2 184. (a) P = 1- 2 t 2 t t 1- ( + 3 2 ) t 3 2 t t 1- t (b)- 2 p + p 1 = 0 2 p- ( + 3 2 ) p 1 + p 2 = 0 3 2 p 1- p 2 = 0 p + p 1 + p 2 = 1 Solving yields p = 2 2 + 2 + 3 2 p 1 = 2 2 + 2 + 3 2 p 2 = 3 2 2 + 2 + 3 2 185. (a) p ii = 1 implies that State 1 is an absorbing state, so States 0 and 3 are...
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chapter18 - 1(a = 2 arrivals/hr = 3 services/hr = 2 3 P...

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