Equivalency of probabilities *In a hospital, the probability that a patient has Disease M is $0.003$. A test is carried out for Disease M. This test has a probability p of giving a positive result if patient has Disease M and the same probability p of giving a negative result if patient does not have Disease M. A patient is given the test.
Taking p to be $0.897$, find the probability that the result of the test is positive.

This problem can be very easily solved with the aid of a tree diagram. What I have a problem with is this:

To find the p(test positive), I simply take $0.003p + 0.997(1-p)$, and of course I'll then use $p = 0.897$. I know this in an extremely mechanistic manner and I feel like I don't truly understand it intuitively. What I'm finding now is p(patient has disease and test positive) + p(patient does not have disease and test positive). Why does this automatically give me p(test positive)? Why are they equivalent? Can somebody explain in a way which would allow me to understand this intuitively? Ridiculous as it may sound, why can't I just say, the probability of the test being positive is simply p + (1-p), i.e. disregarding whether the patient actually has the disease or not? I mean, the question asks for the probability of the test being positive - it didn't ask about the status of the patient!
I'm not sure whether or not this will help you to understand, but I will give it a try...



First, split the given scenario into disjoint events:


The patient is sick and the test is positive
The patient is not sick and the test is positive
The patient is sick and the test is not positive
The patient is not sick and the test is not positive


Then, find the probabilities of the desired events:


The patient is sick $\color\red{\text{and}}$ the test is positive: $(0.003)\color\red\times(0.897)$
The patient is not sick $\color\red{\text{and}}$ the test is positive: $(1-0.003)\color\red\times(1-0.897)$


Finally, find the probability of the unified event:


The 1st event $\color\green{\text{or}}$ the 2nd event: $(0.003)\times(0.897)\color\green+(1-0.003)\times(1-0.897)$
What I'm finding now is p(patient has disease and test positive) +
  p(patient does not have disease and test positive). Why does this
  automatically give me p(test positive)


Let $B$ the event a patient is diseased-$\overline B$ the converse event. 

And $A$ is the event that the test is positive. 

You are asking why 

$=P(A\cap B)+P(A\cap \overline B)=P(A)$

To see that this is true you can draw a venn diagram:



1) I think you can identify  $A \cap B$. 

2) To get $A\cap \overline B$ you have to mark $A$ first. I have marked it with green lines. And $\overline B$ is the area marked with black lines. The intersection of $A$ and $\overline B$ is the area with green and black lines. You can say it is the area $A$ without the intersection of $A$ and $B$.

You have to sum up the areas of 1) and 2) to get $A$. I hope it is comprehensible.



Using the equation of conditional probability $P(A\cap B)+P(A\cap \overline B)$ is equal to 

$P(B)\cdot P(A|B)+P(\overline B)\cdot P(A|\overline B)=P(A)$

with $P(B)=0.003, P(\overline B)=1-P(B)=1-0.003, P(A|B)=0.897$ and 

$P(A|\overline B)=1-P(\overline A|\overline B)=1-0.897$

Thus $P(A)=0.003\cdot 0.897+(1-0.003)\cdot (1-0.897)$