Conditional probability and disease – Mathematics Stack Exchange

Let $Y=1$ denote having a disease. The disease hits 1% of the population. A test gives us the following results
$$P(X=1|Y=1)=0.998$$$$P(X=-1|Y=0)=0.993$$$X=1$ denotes that a test is positive i.e that the disease is in that person and $T=-1$ denotes that the patient is negative in terms of the disease.

  • What is $P(X=1)$ and $P(X=-1)$?
  • Calculate the conditional probability that the patient does not have the disease given a negative test.

My try

For the first one. I have created a 2 times 2 chart but I can not get the columns to equal 1. I would guess that the probability as it is given in the description for having the disease is 1% so $P(X=1)=0.01$ and $P(X=-1)$ would be unkown to me.

For the second one. I Would have to use Bayes theorem for $P(X=-1|Y=0)$. I can not do this with the missing result from the first question.