Probability of child1 given child2 =? 
*

*Let A, B and C be 3 random variables with binary outcomes $\{0,1\}$ defined by the probabilistic graph model below.

*If B is observed as $(B=1)$ & A and C are not observed.

*What is the probability that C = 1 given B = 1?

*We have $P(B|A)$, $P(C|A)$ and $P(A)$
I am searching for the formula only.

 A: I assume the random variables are chosen uniformly?
If that's the case:

"I am searching for the formula only."
Then i guess this is the formula you're after?:
∀ x∈U: P(x) = 1/|U|
the above formula(called the Uniform Distribution)
Means that,

*

*for all/every(∀) element x in(∈) Universe(U):

*the Probability of x is 1 ÷ the size of U (denoted |U|)


A try at the answer itself: Correct me if I'm wrong:
Isn't the probability that C = 1 given B=1,
1/3 or,  0,3?

Am I missing something?
A: Your notation seems to be conflicted about whether $A,B,C$ are binary random variables or events. I'm going to go with events because that makes writing things easier, but of course, these are equivalent.
Also, I am assuming $B$ and $C$ are conditionally independent given $A$, which you've implied but not actually said outright.
In that case, we can figure out $\Pr[A \cap B \cap C]$ as $\Pr[A] \cdot \Pr[B \mid A] \cdot \Pr[C \mid A]$. (Normally, this would have $\Pr[C \mid A \cap B]$, but we are assuming conditional independence here.) We can also figure out $\Pr[A^c \cap B \cap C]$ as $\Pr[A^c] \cdot \Pr[B \mid A^c] \cdot \Pr[C \mid A^c]$.
Finally, $\Pr[B \cap C] = \Pr[A \cap B \cap C] + \Pr[A^c \cap B \cap C]$, and $\Pr[C \mid B] = \frac{\Pr[B \cap C]}{\Pr[B]}$.
