Respuesta :
Answer:
a) [tex]P(B|A) =\frac{P(A and B)}{P(A)}=\frac{0.65}{0.75}=0.867[/tex]
b) In order to two events can be considered as independent we need to have this condition:
[tex]P(A and B) = P(A)*P(B)[/tex]
If we check the condition we see this:
[tex]P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)[/tex]
So since we don't have the condition satisfied the events "do the homework regularly" and "pass the course" are not independent.
Other way to check independence is with the following two conditions:
[tex]P(B|A)=P(B)[/tex] or [tex]P(A|B)=P(A)[/tex], and we see that [tex]P(B|A)\neq P(B)[/tex] for our case.
Step-by-step explanation:
Data given
Lets define the following events
A: Represent the students who do homework regularly
B: Represent the students who pass the course
75% of her students do homework regularly
70% of her students pass the course
65% of her students do homework and also pass the course
For this case we have some probabilities given:
P(A) =0.75, P(B) =0.70 and P(A and B)=0.65
Solution to the problem
Part a
On this case we want to find this probability:
P(B|A) the conditional probability (Pass the course given that he does the homework regularly).
By defintion of conditional probability we know that:
[tex]P(B|A) =\frac{P(A and B)}{P(A)}[/tex]
And now we can replace in the conditional formula like this:
[tex]P(B|A) =\frac{P(A and B)}{P(A)}=\frac{0.65}{0.75}=0.867[/tex]
Part b
In order to two events can be considered as independent we need to have this condition:
[tex]P(A and B) = P(A)*P(B)[/tex]
If we check the condition we see this:
[tex]P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)[/tex]
So since we don't have the condition satisfied the events "do the homework regularly" and "pass the course" are not independent.
Other way to check independence is with the following two conditions:
[tex]P(B|A)=P(B)[/tex] or [tex]P(A|B)=P(A)[/tex], and we see that [tex]P(B|A)\neq P(B)[/tex] for our case.
