Need stats help. prob that patient has cancer given the test was negative

A Canadian male has recently had a Prostate Specific Antigen (PSA) test as to determine if he has prostate cancer. The false-positive rate of a PSA test is 14%. If he does have prostate cancer, PSA test will be positive 81% of the time.

Because this male is showing symptoms that are consistent with prostate cancer, it is assumed that the chance he has prostate cancer prior to taking the PSA test is 0.17.

Part (a) What is the probability that the PSA test will yield a positive result?

Part (b) If the PSA test gives a positive result, what is the probability that he does not have prostate cancer?

Part (c) Suppose the PSA test result is negative, indicating that he does not have prostate cancer and his symptoms are a result of something else. What is the probability that he does have prostate cancer?

I've already solved A and B but for some reason I can't get C right. I have all the numbers i need in a table and am using this equation. P(D|-) = P(Dn-)/P(-) where D is having the disease and - means the result was negative.

First we compute the individual probabilities that the patient has (has not) the cancer and the test is positive (negative): $$P(C\cap+)=P(C)P(+|C)=0.17×0.81=0.1377$$ $$P(C\cap-)=P(C)(1-P(+|C))=0.17×0.19=0.0323$$ $$P(\neg C\cap+)=(1-P(C))P(+|\neg C)=0.83×0.14=0.1162$$ $$P(\neg C\cap-)=(1-P(C))(1-P(+|\neg C))=0.83×0.86=0.7138$$ Then the required probability is $$P(C|-)=\frac{P(C\cap-)}{P(C\cap-)+P(\neg C\cap-)}=\frac{0.0323}{0.0323+0.7138}=0.0433$$
First, let us denote by $$D$$ (disease) the event that this individual has prostate cancer. Let $$T$$ (test positive) denote the event that the patient's test result is positive. Let us also use the notation $$\bar D$$ and $$\bar T$$ to denote the complementary events of not having prostate cancer, and having a negative result, respectively.
Now, we are told that for this individual, $$\Pr[D] = 0.17$$, and for the test being used, we have $$\Pr[T \mid \bar D] = 0.14, \\ \Pr[T \mid D] = 0.81.$$ We are asked to find $$\Pr[D \mid \bar T],$$ the probability the individual actually has prostate cancer, given that the test results were negative. Clearly, we should apply Bayes' theorem since the conditional probabilities we were given are "reversed"; therefore $$\Pr[D \mid \bar T] = \frac{\Pr[\bar T \mid D]\Pr[D]}{\Pr[\bar T]}.$$ Now, we know what $$\Pr[D]$$ is, but we neither know $$\Pr[\bar T \mid D]$$ nor $$\Pr[\bar T]$$. The first is easy to address because it is the complementary probability to $$\Pr[T \mid D]$$, since given that $$D$$ is true, either the test comes back positive or negative (we assume the test does not give an inconclusive result). That is to say, $$\Pr[T \mid D] + \Pr[\bar T \mid D] = 1,$$ so $$\Pr[\bar T \mid D] = 1 - 0.81 = 0.19.$$ To get $$\Pr[\bar T]$$, we have to use the law of total probability: $$\Pr[\bar T] = \Pr[\bar T \mid D]\Pr[D] + \Pr[\bar T \mid \bar D]\Pr[\bar D],$$ where we have conditioned the event $$\bar T$$ on whether $$D$$ or $$\bar D$$ is true--i.e., either the individual does or does not have prostate cancer. Note the first term is the same as the numerator in Bayes' theorem, so that's already done. The second term is found by using the same idea as above and writing it in terms of the complementary probabilities: $$\Pr[\bar T \mid \bar D]\Pr[\bar D] = (1 - \Pr[T \mid \bar D])(1 - \Pr[D]) = (1 - 0.14)(1 - 0.17).$$ Now that you have all the pieces, the computation is straightforward.
If all of this is difficult to navigate, it may help to frame the question in frequentist terms. Construct a $$2 \times 2$$ contingency table. Suppose $$10000$$ people are like this one male. $$1700$$, or $$17\%$$, of them will have prostate cancer (event $$D$$). Of these $$1700$$ who do, $$81\% = 1377$$ of them will test positive (event $$T \cap D$$). Of the $$10000 - 1700 = 8300$$ who do not have prostate cancer (event $$\bar D$$), $$14\% = 1162$$ will experience a false positive result (event $$T \cap \bar D$$). This is summarized as follows: $$\begin{array}{c|c|c|c} & D & \bar D & \\ \hline T & 1377 & 1162 & 2539 \\ \hline \bar T & 323 & 7138 & 7461 \\ \hline & 1700 & 8300 & 10000 \end{array}$$ Note that the totals in the last column sum to $$10000$$, which lets us check that we did the math correctly. Now all we need to do to compute $$\Pr[D \mid \bar T]$$ is to find, out of the $$7461$$ men who test negative (event $$\bar T$$), how many actually had prostate cancer, which is $$323$$ (event $$D \cap \bar T$$). Then the desired probability is $$\Pr[D \mid \bar T] = \frac{\Pr[D \cap \bar T]}{\Pr[\bar T]} = \frac{323}{7461}.$$ Check this against the computations done with the first method.