Respuesta :
Answer:
a) [tex] P(E) = \frac{1}{6}[/tex]
b) [tex] P(F) = \frac{6}{36}=\frac{1}{6}[/tex]
c) [tex]P(E and F) = \frac{1}{36}[/tex]
d) [tex]P(E|F) = \frac{P(E and F)}{P(F)}= \frac{1/36}{1/6}=\frac{1}{6}[/tex]
e) [tex]P(F|E) = \frac{P(E and F)}{P(E)}= \frac{1/36}{1/6}=\frac{1}{6}[/tex]
f) A. yes
[tex] P(E and F ) = P(E)*P(F)[/tex]
Step-by-step explanation:
We are assuming on this case that the two dice are fair. And we have the following events defined:
E:{ A 6 on the blue die } F:{ The numbers are equal }
(a) P(E)=
For this case the probability of obtain a 6 on the blue die is
[tex] P(E) = \frac{1}{6}[/tex]
Since we have 6 possible optiosn and just one with a 6 on the blue die.
(b) P(F)=
First we need to define the sampling space when we tossed the two dice.
S={ (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)}
And as we can see we have just 6 pairs on which we have the same numbers for the two die. (1,1), (2,2),(3,3), (4,4),(5,5),(6,6), and the total outcomes are 36 so then:
[tex] P(F) = \frac{6}{36}=\frac{1}{6}[/tex]
(c) P(E and F)=
For this case we assume that the blue die is the first outcome on the sample space. We want to find on which pairs we have a 6 on the blue die and the numbers are equal in the two dice.
The only pair that satisfy this is (6.6) and we have a total of 36 possible outcomes so then:
[tex]P(E and F) = \frac{1}{36}[/tex]
(d) P(E|F)=
For this case we can use the following formula from the Bayes theorem:
[tex]P(E|F) = \frac{P(E and F)}{P(F)}= \frac{1/36}{1/6}=\frac{1}{6}[/tex]
(e) P(F|E)=
For this case we can use the following formula from the Bayes theorem:
[tex]P(F|E) = \frac{P(E and F)}{P(E)}= \frac{1/36}{1/6}=\frac{1}{6}[/tex]
Are events E and F independent?A. yesB. no
A. yes
Because we have that the following condition satisfied:
[tex] P(E and F ) = P(E)*P(F)[/tex]
