Answer:
a) p(the student did not do homework and he/she passed the course) = 0.09
b) p(the student did not do homework given that she/he passed the course) = 0.125.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
![P(B|A) = \frac{P(A \cap B)}{P(A)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D)
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
a. p(the student did not do homework and he/she passed the course)
If a student does not do homework most days, the chance of passing the course is only 30%. 100 - 70 = 30% don't do homework on a regular basis.
So
0.3*0.3 = 0.09
p(the student did not do homework and he/she passed the course) = 0.09
b. p(the student did not do homework given that she/he passed the course)
Conditional probability.
Event A: Passed the course
Event B: Did not do homework.
p(the student did not do homework and he/she passed the course) = 0.09
This means that ![P(A \cap B) = 0.09](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.09)
Probability that the student passes the course:
90% of 70%(do homework)
30% of 30%(do not do homework).
This means that:
![P(A) = 0.9*0.7 + 0.3*0.3 = 0.72](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.9%2A0.7%20%2B%200.3%2A0.3%20%3D%200.72)
p(the student did not do homework given that she/he passed the course)
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.09}{0.72} = 0.125](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.09%7D%7B0.72%7D%20%3D%200.125)
So
p(the student did not do homework given that she/he passed the course) = 0.125.