conditional probability of continuous independent random variables I would like to know what's the exact way to obtain the conditional probabilities of node having multiple independent continuous random variables as its parent.
Say something like a Noisy OR gate model, with multiple parents of continuous random variable types. 
The samples, I stumbled upon so far (for Noisy OR gate) where explained using discrete binary random variables, but my interest mostly lies on achieving similar outcomes with one or more continuous & random as inputs?
Let me explain with a simple example from Google results. 
A simple study says Bronchitis (B), Tuberculosis (T) and Lung cancer (L) causes Fatigue (F). It also says probability of B -> F = 0.6, T -> F = 0.7,  L -> F = 0.8. with this information if we had to find out P(F|B,T,L), then the way to do this is find out
P(no F|B, T, L) = (1-0.6)(1-0.7)(1-0.8) = 0.024
then P(F|B,T,L) = 1 - P(no F|B, T, L) = 0.976.
In the above case all the causation random variable are of type discrete.
In case if we add a continuous random variable say age to the scenario.
B -> F = 0.6, T -> F = 0.7,  L -> F = 0.8 
and if age (a) is 
        a <= 30 -> F = 0.1, 
   30 > a <= 50 -> F = 0.2, and  
        a >  50 -> F = 0.7. 
Then if I had to find out
P(F|B,T,L,30 > a <= 50), then what should be logic to find it out?
If just look at the example without considering age, we never know the joint probability distribution of of two or more individual events, but the final conditional probability of F given B,L & C has happened at ones is obtained using the individual probabilities of B,L & C. 
Similarly I would like to know the way to compute P(F|B,T,L,30 > a <= 50)? Where 'a' is a discretized continuous random variable to ease the analysis.
Thanks,
Kamal.
 A: (I'm just going to rewrite what I wrote in the comments section above regarding the scenario in your question before I talk about the new scenario you described in the comments for the sake of making this a complete answer to your original post.)
Let's denote the presence of a cause of fatigue by some letter $X$ and the event that that cause leads to fatigue by $X_f$, so that the probability that a cause of fatigue leads to fatigue is $P(X_f)$.  Given the scenario you describe regarding the independent causes of fatigue, $F$, then you're right to say
$$P(F|B,T,L)=1-(1-P(B_f))(1-P(T_f))(1-P(L_f))$$
Now we consider an additional cause to fatigue that is independent of the other causes, in this case age $A$, that has different probabilities of leading to fatigue depending it's value.  Given that $P(A_f|A\lt 30)=0.1, P(A_f|30\lt A\le 50)=0.2$, and $P(A_f|A\gt 50)=0.7$, we can now treat each of these cases as causes themselves that are independent to $B, L,$ and $T$. So, for example
$$P(F|B,T,L,30\lt A\le 50)=1-(1-P(B_f))(1-P(T_f))(1-P(L_f))\left(1-P(A_f|30\lt A\le 50)\right)$$
In the other scenario you describe in the comments above we have the following data (modified slightly so that it adds up):
16 of the first 29 ages lead to $F$, 8 of the next 21 ages lead to $F$, and 17 of the last 50 ages lead to $F$.  If we assume that each patient is equally likely to be any of the ages within their age group then we can write that data like this:
$$P(A_f|A\lt 30)=\frac{16}{29}$$$$P(A_f|30\le A\le 50)=\frac{8}{21}$$$$P(A_f|A\gt 50)=\frac{17}{50}$$
Now this is the same as the original scenario in your question and we'd still be able to use the same general formula, for example:
$$P(F|B,T,L,30\le A\le 50)=1-\left(1-P(B_f)\right)\left(1-P(T_f)\right)\left(1-P(L_f)\right)\left(1-P(A_f|30\le A\le 50)\right)$$
If three patients, one in each age group, came to see a nurse then the probability that one or more of them had fatigue based solely on their age would be:
$$1-\left(1-P(A_f|A\lt 30)\right)\left(1-P(A_f|30\le A\le 50)\right)\left(1-P(A_f|A\gt 50)\right)=$$
$$1-\left(1-\frac{16}{29}\right)\left(1-\frac{8}{21}\right)\left(1-\frac{17}{50}\right)\approx0.817$$
Let's say that $A$ isn't a set of discrete points, though, and instead say that it's modeled by a continuous distribution.  Then we'd have a probability density function, $f_A(a)$, such that $P(A_f|s\lt A\lt t)=\int_s^t{f_A(a)}da$ and we'd still be able to use the same general formula as we did in the discrete case:
$$P(F|B,T,L,30\le A\le 50)=1-\left(1-P(B_f)\right)\left(1-P(T_f)\right)\left(1-P(L_f)\right)\left(1-\int_{30}^{50}f_A(a)da\right)$$
