Applied Probability- Bayes theorem I need help in all things related to identifying, defining conditions and solution feed back and reasoning most importantly.
1) A blood test indicates the presence of a particular disease 95 % of the time when the diseases is actually present. The same test indicates the presence of the disease .5% of the time when the disease is not present. One percent of the population actually has the disease. Calculate the probability that a person has a disease given that the test indicates the presence of the disease.
First and foremost one might think that because the word "given" is followed by the indication of the disease they might think that, the indication of the disease is the is the condition However my understanding is that a condition related to a mathematical definition is the state of affairs that must occur or exist before something else can happen. Well clearly one must first either have the disease or not have the disease before a test can indicate whether or not the disease is in fact present. 
Solution:
$C=$ The individual has the disease $.01%$
$E=$ The test indicates that it is present in those with the disease $.95%$
SO we want the proportion of those that have the disease and test positive to the proportion of those that have the disease:
$P(E \vert C) = \frac{P(E \cap C)}{P(E \cap C)+P(E^c \cap C)}$
$= \frac{(.01)(.95)}{(.01)(.95)+(.05)(.01)}= \frac{.0095}{.01}=.950$
My answer choices were $.324,.657,.945,.950,.995$
This one made the most sense to me? The other method I used by using the condition as the test didnt even give me anything close to 1%. Was I supposed to do it the other way and was I somehow supposed to take into account that of those that don't have the disease 50% test positive but that wouldnt make sense to me. 
2) Ninety-eight percent of all babies survive delivery, However 15 percent of all births involve a c section and when they are done the baby survives 96 percent of the time. If a randomly chosen pregnant woman does not have a c section what is the probability that her baby survives?
By the same logic as the previous problem the condition would only make sense to be the method of delivery because if we are talking about a baby surviving delivery then its survival is conditioned on the method of delivery. Correct? 
$C=$ Does not have a C-section $.85%$
$E=$ her baby survives without a c-section this is difficult to figure out:
if 96% of all 15% c section births survive and since we have the percentage of the total 98 percent that survive from both non c and c section then thus 1 minus this should give us the percentage that survive when they do not have a C-section, this makes sense correct? then the probability of a baby surviving a c-section is $.98-.144=.836$ Right?
Okay now we set up the question:
Using this we want:
$P(E \vert C)= \frac{P(EC)}{P(E \cap C)+ P(E^c \cap C)}= \frac{P(E \cap C)}{P(C)}=\frac{(.836)(.85)}{(.836)(.85)+ ?}$
?= we want the percentage that didn't survive when they had a c-section and didn't have a c-section $(.85)(1-.836?)$
recopying and pasting we get:
$P(E \vert C)= \frac{P(EC)}{P(E \cap C)+ P(E^c \cap C)}= \frac{P(E \cap C)}{P(C)}=\frac{(.836)(.85)}{(.836)(.85)+(1-.836)}=\frac{.7106}{(.7106+.164)}$
Some how this was wrong as the supposed answer is slightly above .98. I want to rant a bunch of improper things and swear at the top of my lungs who the hell does my logic not follow I used the exact definition of condition and and properly applied bayes theoreom please label every step and the reason behind each intersection obtained.
 A: 1)
Define:


*

*$P(C)$ = prior probability a patient has the disease:  $P(C) = 0.01$

*$P({\bar C})$ = prior probability a patient does not have the disease:  $P({\bar C}) = 1 - 0.01 = .99$

*$P(E|C)$ = probability of a positive test result given the patient has the disease:  $P(E|C) = 0.95$

*$P(E| {\bar C})$ = probability a positive test result given the patient does not have the disease:  $P(E| {\bar C}) = 0.005$  


You want $P(C|E)$ = the probability the patient has the disease given the test yields a positive result.
Use Bayes rule:
$P(C|E) = {P(E|C) P(C) \over P(E)} = {P(E|C) P(C) \over P(E|C)P(C) + P(E|{\bar C}) P({\bar C})}$.
Use this Bayesian approach for 2)
A: I find the easiest way to do these problems is with a contingency table like this one. Pick a large population so you can deal with whole numbers rather than percents. I used 100,000 here:
             has disease   healthy    total
tests +          950           495     1445
tests -           50         98505    98555
total           1000         99000   100000

Now the probability that someone who tests positive actually has the disease is easy to calculate given the numbers in the first row:
$$
\frac{950}{1445} = 0.65743944636 \approx 0.657 .
$$
That answer is much too precise. The data have just one or two significant digits so it makes no sense to report anything more precise than $0.66$ or even $0.7$. In fact "about $2/3$" would be the most informative.
This is in fact just Bayes' theorem, but with less chance to make a mistake and (I think) more understanding.
Can you set up the table for the second problem?
References:
Chances Are:
https://opinionator.blogs.nytimes.com/2010/04/25/chances-are/
Natural frequencies improve Bayesian reasoning in simple and complex inference tasks
http://journal.frontiersin.org/article/10.3389/fpsyg.2015.01473/full
