Question 589469
Construct a table:

             
---------Know----Doesn't Know
Correct   1-----------1/5
Incorrect 0-----------4/5


Our sample space is limited to:


the student knew the answer and got it correct
the student did not know the answer but got it correct.


The first case is easy, P(know) = 3/4 and P(correct) = 1, so P(know and correct) = 3/4.


For the second case, P(doesn't know) = 1/4 and P(correct) = 1/5 (since there are 5 choices, one selected at random). Therefore P(doesn't know and correct) = (1/4)(1/5) = 1/20.


The probability that he got it correct is 3/4 + 1/20 = 13/20. However, we are only interested in P(know and correct) which is 3/4, or 12/20. Therefore, P(know | correct) = 12/13.