Respuesta :
Answer:
Step-by-step explanation:
The sample space consists of the 36 elementary events, i.e., n(S)=36.
A be the event that the red die shows 3 dots is (3,1),(3,2),(3,3),(3,4),(3,5),(3,6). Then P(A)=n(A)/n(S)=(6/36)=(1/6).
B be the event that the green die shows 4 dots is (1,4),(2,4),(3,4),(4,4),(5,4),(6,4). Then P(B)=n(B)/n(S)=(6/36)=(1/6).
C be the event that the total number of dots showing on the 2 dice is 7 is (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). Then P(C)=n(C)/n(S)=(6/36)=(1/6).
Now (A\capB)=(3,4) => P(A\capB)=(1/36) , (A\capC)=(3,4) => P(A\capC)=(1/36), (B\capC)=(3,4) => P(B\capC)=(1/36) & (A\capB\capC)=(3,4) => P(A\capB\capC)=(1/36).
Consider P(A\capB)=P(A)P(B)=(1/6)(1/6)=(1/36), then A and B are pairwise independent.
P(A\capC)=P(A)P(C)=(1/6)(1/6)=(1/36), then A and C are pairwise independent.
P(B\capC)=P(B)P(C)=(1/6)(1/6)=(1/36), then B and C are pairwise independent.
Therefore, these events A, B & C are pairwise independent.
Now consider P(A\capB\capC)=P(A)P(B)P(C)=(1/6)(1/6)(1/6)=(1/216) not = (1/36).
Therefore, A,B & C are not mutually independent events.
Answer:
The events A, B and C are pairwise independent but not mutually independent.
Step-by-step explanation:
The sample space when you roll two dice is given below.
(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)
From the above, Total Number of Sample Space=36
A is the event that the Red die shows 3 dots.
A=(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
n(A)=6
B is the event that the green die shows 4 dots.
B=(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
n(B)=6
C is the event that the total number of dots showing on the two dice is 7.
C= (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
n(C)=6
Definition
Events A, B and C are pairwise independent if the following holds.
P(A∩B)=P(A)P(B)
P(A∩C)=P(A)P(C)
P(B∩C)=P(B)P(C)
A∩B ={(3,4)}
P(A∩B)= [TeX]\frac{1}{36}[/TeX]
P(A)=[TeX]\frac{6}{36}[/TeX]
P(B)=[TeX]\frac{6}{36}[/TeX]
P(A)P(B)=[TeX]\frac{6}{36} X \frac{6}{36} = \frac{1}{36} [/TeX]
A∩C={(3,4)}
P(A∩C)= [TeX]\frac{1}{36}[/TeX]
P(A)=[TeX]\frac{6}{36}[/TeX]
P(C)=[TeX]\frac{6}{36}[/TeX]
P(A)P(B)=[TeX]\frac{6}{36} X \frac{6}{36} = \frac{1}{36} [/TeX]
B∩C={(3,4)}
P(B∩C)= [TeX]\frac{1}{36}[/TeX]
P(B)=[TeX]\frac{6}{36}[/TeX]
P(C)=[TeX]\frac{6}{36}[/TeX]
P(B)P(C)=[TeX]\frac{6}{36} X \frac{6}{36} = \frac{1}{36} [/TeX]
Since the conditions below are true
P(A∩B)=P(A)P(B)
P(A∩C)=P(A)P(C)
P(B∩C)=P(B)P(C)
The sets A, B and C are pairwise independent.
DEFINITION
The sets A, B and C are mutually independent if:
P(A∩B∩C) = P(A)P(B)P(C)
A∩B∩C = {(3,4)}
P(A∩B∩C) = [TeX]\frac{1}{36} [/TeX]
P(A)P(B)P(C) = [TeX]\frac{1}{6} X \frac{1}{6} X \frac{1}{6}= \frac{1}{216}[/TeX]
Since P(A∩B∩C) ≠ P(A)P(B)P(C), the events bare not mutually independent.
