ProblemasExamenes

# ProblemasExamenes - Ejercicios de exmenes Francisco Parreo...

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Ejercicios de exámenes. Francisco Parreño Torres. Probabilidades de transición p 00 = P{( ¬ averia) (1 averia repara) (1 averia repara)} = = (0.9 × 0.9) + (0.1 × 0.9 × 0.6) + (0.1 × 0.9 × 0.6) = 0.918 p 01 = P{(1 averia ¬ repara) (1 averia ¬ repara)} = = (0.1 × 0.9 × 0.4) + (0.1 × 0.9 × 0.4) = 0.072 p 02 = P{(1 averia 1 averia)} = (0.1 × 0.1) = 0.01 p 03 = 0 p 10 = P{( ¬ averia repara)} = (0.9 × 0.6) = 0.54 p 11 = P{( ¬ averia ¬ repara) (1 averia repara)} = (0.9 × 0.4) + (0.1 × 0.6) = 0.42 p 12 = P{(1 averia ¬ repara)} = (0.1 × 0.4) = 0.04 p 13 = 0 p 20 = 0; p 21 = 0; p 22 = 0; p 23 = 1 p 30 = 1; p 31 = 0; p 32 = 0; p 33 = 0 Matriz de probabilidades de transición = 0 0 0 1 1 0 0 0 0 04 . 0 42 . 0 54 . 0 0 01 . 0 072 . 0 918 . 0 P Diagrama de transiciones 1 3 2 0 0.918 0.072 0.54 0.01 1 1 0.42 0.04 b) Clasificar los estados de la cadena y hallar las clases de equivalencia. Todos los estados son recurrentes positivos y aperiódicos. Hay una única clase de equivalencia formada por todos los estados {0,1,2,3,4}.
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ProblemasExamenes - Ejercicios de exmenes Francisco Parreo...

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