Conditional Probabilty Question 
"John is taking an algebra and a calculus class. The probability that
  he will pass algebra is 0.57 and the probability he passes calculus is
  0.68. The probability he passes calculus given he passes algebra is 0.72, and the probability that he passes calculus given he doesn't pass algebra is 0.45. What is the probability he passes algebra given
  he passes calculus."

My approach:
$$P(C|A)\cdot P(A) = P(A \cap C)$$
Then $\dfrac{P(A\cap C)}{P(C)} = 0.6035$, 
but this is incorrect according to the professor, and that the answer should be 0.6796   
Can anyone explain this? Thank you in advance!
 A: Below are the detailed calculation. Then reason why I provide all the details is because it seems to be a mistake in the statement.

Apply Bayes' formula to obtain
$$
\Pr(A \,|\, C)
= \frac{\Pr(C \,|\, A) \Pr(A)}{\Pr(C)}
$$
The numerator is $0.72 * 0.57 = 0.4104$.
The denominator can be calculated using the law of total probability
$$
\Pr(C) = \Pr(C \,|\, A) \Pr(A) + \Pr(C \,|\, \text{not }A) \Pr(\text{not }A)
$$
$$
= 0.72 * 0.57 + 0.45 * 0.43 = 0.6039
$$
Thus,
$$
\Pr(A \,|\, C) = \frac{0.4104}{0.6039} = 0.6796   
\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } 
\text{(not 0.6769)}
$$

Note that the statement says that $\Pr(C) = 0.68$, which, according to the above calculation, is wrong.
A: I think the critical factor is to ignore the given that the probability he passes calculus is 68%.  This is actually not correct given the other information in the problem.
If the probability that he passes algebra is 57%, the probability that he passes calculus given passing algebra is 72%, and the probability that he passes calculus while failing algebra is 45%, then
$$P(C) = P(C|A)P(A) + P(C|A')P(A') = 0.57*0.72 + (1-0.57)*0.45 \approx 0.604$$
Dividing $P(C \cup A)$ by 0.6039 instead of 0.68 will give you the 0.68 the prof is looking for.
A: Expand the denominator using Law of Total Probability and the correct answer follows.
Your mistake was to write $P(C|A)\cdot P(A) = P(A \cap C)$ since this is also equal to $P(A|C)\cdot P(C)P(A|C)\cdot P(C)$ where $P(A|C)P(A|C)$ is the quantity we want to find, so this would become a self-referential problem and hence unsolvable this way.
