TAREA 4 ESTADISTICA
Enviado por jorge.varelacu • 16 de Diciembre de 2012 • 1.034 Palabras (5 Páginas) • 1.829 Visitas
Escriba en cada caso presentado, el espacio muestral correspondiente y luego calcule las probabilidades de cada evento.
1. Se lanzan dos dados de seis caras, construya el espacio muestral asociado y calcule la probabilidad que existe de que ocurran los siguientes eventos o sucesos:
Espacio muestral:
1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6
2.1 – 2.2 – 2.3 – 2.4 – 2.5 – 2.6
3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6
4.1 – 4.2 – 4.3 – 4.4 – 4.5 – 4.6
5.1 – 5.2 – 5.3 – 5.4 – 5.5 – 5.6
6.1 – 6.2 – 6.3 – 6.4 – 6.5 – 6.6
a) A = {que salgan dos pares}
Posibles combinaciones:
22 - 24 – 26 - 42 – 44 – 46 - 62 - 64 - 66 =>
P = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) +
(1/6)(1/6) + (1/6)(1/6) =>
P = 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 =>
P = 9/36 = 1/4 = 0.25 = 25%
b) B = {el primero sea mayor que el segundo}
Posibles combinaciones:
21 31 32 41 42 43 51 52 53 54 61 62 63 64 65 =>
P = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) +
(1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) =>
P = 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 =>
P = 15/36 = 5/12 = 0.4167 = 41.67%
c) E = {Ambos pares o distintos}
Posibles combinaciones para ambos pares (calculado en el problema A):
22 – 24 – 26 – 42 – 62 – 44 – 46 – 64 - 66 => P = 1/4 = 0.25 = 25%
Posibles combinaciones para distintos:
12 – 13 – 14 – 15 – 16 - 21 – 23 – 24 – 25 – 26 – 31 – 32 – 34 – 35 – 36 – 41 – 42 – 43 – 45
46 – 51 – 52 – 53 – 54 – 56 – 61 – 62 – 63 – 64 - 65 =>
Eliminamos ambos pares, porque ya están calculados previamente:
12 – 13 – 14 – 15 – 16 – 21 – 23 – 25 – 31 – 32 – 34 – 35 – 36 – 41 – 43 – 45 – 51 – 52 – 53 –
54 - 56 - 61 – 62 – 63 – 64 - 65 =>
P₁ = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) +
(1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) +
(1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) +
(1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) =>
P₁ = 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 +
1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36
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