TAREA 4 ESTADISTICA

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 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 +

1/36 =>

P₁ = 26/36 = 13/18 = 0.7222 = 72.22%

Entonces la probabilidad total Pt de este caso es:

Pt = P + P₁ =>

Pt = 1/4 + 13/18 = 9/36 + 26/36 = 35/36 = 0.9722 = 97.22%

2. Se lanzan tres monedas al aire. ¿Cuál es el espacio muestral? ¿Cuál es la probabilidad de?:

Espacio muestral:

c – c – c

c – c – s

c – s – c

c – s – s

s – s – c

s – c – s

s – c – c

s – s – s

a) A = {que sean iguales}

Esto puede ocurrir de dos formas

CCC ó SSS =>

P = (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2) =>

P = 1/8 + 1/8 =>

P = 2/8 = 1/4 = 0.25 = 25%

b) B = {que sean distintas}

Esto puede ocurrir de varias

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