Conditional probability question; Bayes' theory and medical test I have to resolve this conditional probability problems

Observational studies show that in a population of 1% of individuals is carrying a disease. It is available a screening test that it's correct, so positive, for the sick and negative for the healthy, in both directions in 90% of cases. Compute:

*

*(A.) The probability that an individual tests positive;

*(B.) The probability that an individual who is ill will test positive;

*(C.) The probability that an individual who tests positive in two independent trials is sick.


I have set the events $T^+$ "correct test" $D^+$ "sick" and of course the complementary events.
I have used the formula
$\def\P{\mathop{\sf P}} \P(T^+)=\P(T^+\mid D^+)\P(D^+)+\P(T^+\mid D^-)\P(D^-)$ to calculate the first probability [$\P(D^-)=1-\P(D^+)$ probability individual is healthy]
But I can't go through that
Someone could help me?
 A: 
  
*
  
*(A.) The probability that an individual tests positive;
  

$\def\P{\mathop{\sf P}} \P(T^+)=\P(T^+\mid D^+)\P(D^+)+\P(T^+\mid D^-)\P(D^-)$ is indeed the way to go.
That gives you $\P(T^+) ~=~ 0.90\cdot0.01+(1-0.90)\cdot(1-0.01)~=~0.108$

  
*
  
*(B.) The probability that an individual who is ill will test positive;
  

You seek $\P(T^+\mid D^+)$, but you already know what that is.
Did you mean "The probability that an individual who tests positive is ill?"  $\P(D^+\mid T^+)$.   You may use Bayes' Theorem to find this.
Can you use Bayes' Theorem?

  
*
  
*(C.) The probability that an individual who tests positive in two independent trials is sick.
  

Subscript the trials, Find $\P(D^+\mid T_1^+\cap T_2^+)$ assuming that the test accuracy is conditionally independent given any particular patient. $$\P(T_1^+\cap T_2^+\mid D^+) = \P(T_1^+\mid D^+)\P(T_2^+\mid D^+)$$
A: I prefer that we look at this more intuitively than using the often harder to comprehend notation that you used in your question. For these problems in conditional probability, we can just use case work. 
A. In the case that an individual is healthy which happens $99$% of the time, $10$% of those individuals will test positive. In the case that an individual is sick which happens $1$% of the time, $90$% of these individuals will test positive. Thus, the probability of an individual testing positive is $.01*.90+.99*.10 = .108$. Thus, our answer is $10.8$%.
B.) The probability that a person will test positive given that they are sick is stated in the problem as $90$%, which is our answer.
C.) C. If they have tested positive in two trials, we can split up the cases into if they are a sick individual or a healthy individual. In the case of a sick individual which will happen $1$% of the time, they will have a $90$%$*90$% chance of testing positive twice. The probability of this happening is thus $.01*.9*.9=.0081$, so $.81$%.
In the case of a healthy individual, which will happen $99$% of the time, they will have a $10$%$*10$% chance of testing positive in both cases. The probability of this happening is thus $.99*.1*.1=.0099$, so $.99$%.
Now we know that the chance of an individual testing positive in both cases is $.0081+.0099=.018$, and the chance of an sick person testing positive in both cases is $.0081$, so the probability that the person testing positive is sick is simply $.0081/.018=.45$, so our answer is $45$%.
Hope this helps!
