Question

Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final. You randomly pick up a final exam and notice the student received an A. What is the probability that this student earned an A on the midterm?The end-goal is to find P(midterm = A|final = A).

Answer 1

There is a 38.97% probability that this student earned an A on the midterm.

Explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

`P_(A) = 0.13(0.47) + 0.87(0.11) = 0.1568`

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

`P_(AA) = 0.13(0.47) = 0.0611`

The probability that the student also earned an A on the midterm is

`P = (P_(AA))/(P_(A)) = (0.0611)/(0.1568) = 0.3897`

There is a 38.97% probability that this student earned an A on the midterm.