TALLER DE ESTADISTICA II PROBABILIDADES
Enviado por Rebecca • 18 de Diciembre de 2017 • 2.314 Palabras (10 Páginas) • 505 Visitas
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d. oros o un 6
P AoB = P (A) + P (B) – P (A n B)
= 10/40 + 4/40 - 1/40
= 13/40 = 0.325
e. seis de espadas o figura
P AoB = P(A) + P (B) = 12/40 + 1/40 = 13/40
f. una as o figura
P AoB = P (A) + P (B)
= 4/40 + 12/40 = 16/40 = 8/20 =0.4 =40%
2. 20 bolas de plástico: 5 amarillas, 8 negras, 7 rojas.
a. sea negra:
P(A) = 8/20 = 4/10 = 2/5 = 0.4 = 40%
b. No sea amarilla:
P AoB = P(A) + P (B)
= 8/20 + 7/20
= 15/20 = 3/4 =0,75 = 75%
c. sea roja:
P(A) = 7/20 = 0,35 = 35%
d. sea amarilla o negra:
P AoB = P (A) + P (B)
= 5/20 + 8/20
= 13/20 = 0, 65 = 65%
3. P(A) = 0.20; P (B) = 0.70; P(A Y B) = 0.10
a) AyB son mutuamente excluyentes? = No AyB tienen algunos elementos en común, luego no son mutuamente excluyentes.
b) P (AoB) = P (A) + P (B) – P (AyB)
= 0.20 + 0.70 – 0.10 = 0.8
c) P (A’) = 1 – P (A)
P (A’) = 1 – 0.20 = 0.80
d) Son AyB colectivamente exhaustivos? NO
4. P AoB = P (A) + P (B) – P (AnB)
= 13/52 + 4/52 – 1/52
= 16/52 = 8/26 = 4/13 = 0.30
5. & = (1, 2, 3, 4, 5,6)
E1 = (1, 3, 5)
P(A) = 3/6 = ½ =0.5 = 50%
E2 = (2, 4, 6)
P(A) = 3/6 = ½ = 0.5 = 50%
Impar o divisible por dos?
P (AuB) = P(A) + P(B) – P (AnB) E1= “impar” 3
= 5/6 + 3/6 E2= “divisible por dos” 3
= 6/6 = 1 E1 n E2 = 0
Par o divisible por 3
P (AuB) = P(A) + P(B) – P (AnB) E1= “par” 3
= 3/6 + 2/6 - 1/6 E2= “divisible por tres” 2
= 4/6 = 2/3 = 0,66 E1 n E2 = 1
6. A (helado 0.7); B (kumis 0.40)
P (A) = 0.7; P (B) = 0.4; P (A n B) = 0.3
P AoB = P (A) + P (B) – P (A n B)
= 0.7 + 0.4 – 0.3 = 0.8
7) A (Sol); B (que llueva)
P (A) = 0.60; P (B) = 0.20; P (A y B) = 0.03
P (A u B) = P(A) + P (B) – P (A u B)
= 0.60 + 0.20 – 0.30 = 0.77
9. a) Azul:
P (A) = 12/30 = 0.40 = 40%
b) A (azul) o B (Amarillo)
P AoB = P(A) + P (B)
= 12/30 + 8/30 =20/30 =10/15 = 0.66
C) A (azul), B (verde)
P AoB = P(A) + P (B)
= 8/30 +10/30 = 18/30 = 3/5 = 0.6
10. A (economistas); B (administradores)
P AoB = P(A) + P (B)
= 6/16 + 4/16
= 10/16 = 5/8 = 0.625
B) SUCESOS INDEPENDIENTES (regla de multiplicación)
1. Baraja de 52 cartas con reposición
a) Al sacar dos cartas ambas sean diamantes
P1 = 13/52 P2 = 13/52
P = P1 * P2
P = 13/52 * 13/52
= ¼ * ¼
= 1/16
b) Ambas sean figuras ( J, Q, K)?
P1 = 12/52 = 3/13 P2 = 12/52 = 3/13
P = P1 * P2
= 3/13 * 3/13
= 9/169
C) corazon y diamante
P1 = 13/52 = ¼ P2 = 13/52 = ¼
P = P1 * P2
= 1/4 * 1/4
= 1/16
2. Del ejercicio anterior: las dos cartas se extraen sin reposicion.
a) ambas sean diamantes
P1 = 13/52 = ¼ P2= 12/51
P = P1 * P2
= 1/4 * 12/51
= 3/51
b) sean ambos figuras ( J, Q, K)
P1 = 12/52 = 3/13 P2= 11/51
P = P1 * P2
= 3/13 * 11/51
= 11/221
c) corazon y diamante
P1 = 13/52 = ¼ P2= 13/51
P = P1 * P2
=
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