Percentage of risk, Is my calculation correct? X is the worst scenario. In order for X to occur, two things must happen, in sequence.


*

*An event with an occurrence chance on 1/120 must occur. This is EVENT A. 

*if EVENT A occurs, another event with a 2/100 occurrence chance must occur. This is EVENT B.

*EVENT A + EVENT B occurring are X = the worst possible scenario.


So, at the starting point. What are the chances of "X" to occur? I calculated it to be 0.01% percent. Am I correct?
Based on Comments:
following bloodwork, a statistical model used by OBGYN's determines that a fetus has 1/120 chance of having a certain disorder. If this is indeed the case, a certain test can detect it. But that test fails in 2% of the cases (failure=existing disorder not detected). What are the chances of that fetus to be born sick, if the test is taken and upon detection of the disorder he's aborted.
 A: Ok.
I am interpreting the question as follows:
Given the blood work, we have that the chances of a disorder(X) are 1/120.
Given the blood work, if there is no disorder, the test will say deliver.
Given the blood work, if there is disorder, the test says deliver 2% of the time.
(the above two sentences are where most of the ambiguity lies and depends on how the clinical trials for the test were done)
i.e
If (assuming bloodwork)
X = Disorder exists.
T = Test says deliver.
$P(X) = 1/120$ 
$P(T | X) = 0.02$ 
$P(T | \neg X) = 1$
We are interested in $P(X|T) = P(T|X) P(X)/P(T)$
Now $P(T) = P(T|X) P(X) + P(T|\neg X) P(\neg X) = 0.02/120 + 1*(1-1/120) = 0.9918... $
Thus the probability we seek is $0.02 /(120*0.9918) \sim 0.000168$
A: In order to give an answer, we need to use some technical terminology. But conveniently, in this case, the meaning of the technical terms is not very far from their everyday meaning.
If given the information that $A$ has happened, or equivalently here, is true, the probability that $B$ happens is $2/100$, then the probability of $X$, the worst scenario, is indeed
$$\frac{1}{120}\times \frac{2}{100}.$$
This product is $1/6000$, which is not far from your answer.  It actually turns out to be roughly $0.000167$. You rounded down; rounding up gives a result closer to the truth. I would probably use something like $0.00017$.
If $A$ and $B$ are independent, again the probability of the worst possible scenario is the product.  Here independent has a strictly defined technical meaning, but it is fairly close to the informal meaning of "independent."
But the situation can be more complicated.  Suppose that $A$ is lightning strike, and $B$ is fire.  Then $A$ and $B$ are not independent (lightning strikes can cause fires).  In that case, we would have to know something more about the degree of dependence/independence between lightning strikes and fires to find the probability of the worst case.  Certainly that probability would be greater than the simple product $1/6000$. 
Added: The added information clears things up a lot.  We are in the first case that I discussed. The probability that the disease is present is estimated at $1/120$. Given that the disease is present, the probability of non-detection is said to be $2/100$.  Then the calculation that gives probability that the disease is present and remains undetected is correct, the answer is indeed $1/6000$.
A few cautions, however.  There is no reason to trust fully the estimates $1/120$ and $2/100$.  Estimates are usually based on whole populations, and individual factors may make the general probability estimate not accurate. The numbers $1/120$ and $2/100$ are suspiciously simple-looking, they are undoubtedly rough approximations.  Also, medical improvements tend to lower probabilities over time.  And the $2/100$ non-detection rate does not necessarily reflect the probability under best practices.  
