Prevencion

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:

a) A = { que salgan dos pares }

b) B = { el primero sea mayor que el segundo }

c) E = { Ambos pares o distintos }

RESPUESTA:

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) {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%

E) {ambos pares o distintos}

posibles combinaciones para ambos pares:

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...?:

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