Suppose that 60% of the students do homework regularly. It is also known that 80% of students who had been doing homework regularly, end up doing well in the course. Only 20% of students who had not been doing homework regularly end up doing well in the course. Given that a student did well in the course, what is the probability that the student had been doing homework regularly?
A) 0.286
B) 0.857
C) 0.143
D) 0.429

Question

Quizplus · Accepted Answer

Let A be the event that the student did homework regularly, and B be the event that the student did well in the course. We want to find P(A|B), the probability that the student did homework regularly, given that they did well in the course.

We can use Bayes' theorem to solve this:
P(A|B) = P(B|A) * P(A) / P(B)

We know that P(A) = 0.6, P(B|A) = 0.8, and P(B|not A) = 0.2 (since only 20% of students who did not do homework regularly did well in the course). We can use the law of total probability to find P(B):
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
P(B) = 0.8 * 0.6 + 0.2 * 0.4
P(B) = 0.56

Now we can plug in these values into Bayes' theorem:
P(A|B) = 0.8 * 0.6 / 0.56
P(A|B) = 0.857

Therefore, the probability that the student did homework regularly, given that they did well in the course, is 0.857 or approximately 0.86. So the answer is B.

Suppose That 60% of the Students Do Homework Regularly