bayes rule question regarding exam performance related In an exam, there is a problem that $60\%$ of students know the correct answer. However, there is $15\%$ chance that a student picked the wrong answer even if he/she knows it and there is also a $25\%$ chance that a student does not know the answer but guessed it correctly. If a student did get the problem right, what is the chance that this student really knows the answer?
Hint

A = knows answer, B = answers correctly

$P(A\cap B)$ = $0.6\times 0.85$  

$P(A^c\cap B)$ = $0.4\times 0.25$ 
$$P(A|B) = \frac{(0.6\times 0.85)}{(0.6\times 0.85) + (0.4\times 0.25)} = 0.8361$$
In an exam, there is a problem that 60% of students know the correct answer. However, there is 15% chance that a student picked the wrong answer even if he/she knows it and there is also a 25% chance that a student does not know the answer but guessed it correctly. If a student did get the problem right, what is the chance that this student really knows the answer?