Question

Attendance has never been a component of the grade for this course. However, over the years I have noticed that 40% of those who attend regularly receive an above average grade (B or better), while only 10% of those who do not attend regularly receive an above average grade. About 70% of the students attend class regularly. One student is randomly selected at the end of the semester. Find the probability that the student did not attend class regularly given that (s)he did not receive an above average grade

Answer 1

0.3913 = 39.13% probability that the student did not attend class regularly given that (s)he did not receive an above average grade

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Did not receive an above average grade.

Event B: Did not attend class regularly.

Probability of an student not receiving an above average grade:

100 - 40 = 60% of 70%(attend class regularly).

100 - 10 = 90% of 100 - 70 = 30%(do not attend class regularly).

So

`P(A) = 0.6*0.7 + 0.9*0.3 = 0.69`

Did not receive an above average grade and did not attend class regularly:

90% of 30%. So

`P(A \cap B) = 0.9*0.3 = 0.27`

Find the probability that the student did not attend class regularly given that (s)he did not receive an above average grade

`P(B|A) = (P(A \cap B))/(P(A)) = (0.27)/(0.69) = 0.3913`

0.3913 = 39.13% probability that the student did not attend class regularly given that (s)he did not receive an above average grade