Question:

Suppose a student is about to take a multiple choice test has only learned 60% of the material covered in the exam. thus, there is a 60% chance that she will know the answer to the question. However, if she does not know the answer to a question, she still has a 20% chance of getting the right answer by guessing.

a) if we choose a question at random from the exam, what is the probability that she will get it right?

20% of 40 = 8

60% + 8% = 68% chance of getting it right

b) If we know that she correctly answered a question in the exam, what is the probability that she learned the material in the question.

this is where I have trouble, my thoughts were to do it the following. Because we are told that she is correct we are only concerned about when she gets a question right (part a), so.

60/68 * 100 = 88.24%

It has been suggested that this is not the right way to go about it, instead.

Let C be the event "she got it right", G be the event "she guessed correctly" and K be the event "she knows the material".

Since the question states she got it right, P(C)=1

The probability for guessing an answer correctly is P(G)=0.2

So, if she got it right, P(K)=P(C)-P(G)

thus, P(K)=1-0.2

P(K)=0.8

Can someone show me which method is correct and more importantly, why the other method is incorrect. thanks for you time and assistance.