Question

A professor has been teaching accounting for over 20 years. From her experience she knows that 75% of her students do homework regularly. 70% of her students pass the course and 65% of her students do homework and also pass the course. If a student is randomly selected from her class, what is the probability that the student would: a. Pass the course given that he does the homework regularly? b. Are"do the homework regularly" and "pass the course" independent? Why or why not?

Answer 1

Answer:

a) $P(B|A) =\frac{P(A and B)}{P(A)}=\frac{0.65}{0.75}=0.867$

b) In order to two events can be considered as independent we need to have this condition:

$P(A and B) = P(A)*P(B)$

If we check the condition we see this:

$P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)$

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:

$P(B|A)=P(B)$ or $P(A|B)=P(A)$, and we see that $P(B|A)\neq P(B)$  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:

$P(B|A) =\frac{P(A and B)}{P(A)}$

And now we can replace in the conditional formula like this:

$P(B|A) =\frac{P(A and B)}{P(A)}=\frac{0.65}{0.75}=0.867$

Part b

In order to two events can be considered as independent we need to have this condition:

$P(A and B) = P(A)*P(B)$

If we check the condition we see this:

$P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)$

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:

$P(B|A)=P(B)$ or $P(A|B)=P(A)$, and we see that $P(B|A)\neq P(B)$  for our case.