How to calculate the probability when there is uncertainty Lets say:


*

*I have a bag with 100 balls.

*I know 40 are reds and 30 blues.

*And the remainder 30 can (must) be either be blue or red  (and can take no other color). And are very likely to follow the same distribution of the already know set
How the I get (compute) the probability of taking 1 red ball of the bag?

This comes from a real world problem I have (computer science) where sometime I get the color of the ball and other-times it is not given to me (of course the real data is not about the color of the balls, but this anecdote capture the essence of my problem). 
 A: EDIT: The current state of the question is significantly different from the one I tried to answer with this response. Here is the question this response attempts to answer:

Lets say:
  
  
*
  
*I have a bag with 100 balls.
  
*I know 40 are reds and 30 blues.
  
*And the remainder 30 can (must) be either be blue or red  (and can take no other color).
How the I get (compute) the probability of taking 1 red ball of the
  bag?

We can rephrase the question as trying to find the expected probability of picking a red ball out of the bag.
For this, we use a formula of expectation: $$E(X) = \sum_{s \in S} p(s) \cdot x(s)$$ Where $E(X)$ is the expected outcome on a given pick, $S$ is the sample space, $p(s)$ is the probability of this event occurring, and $x(s)$ is the value of the event occurring.
You have to consider every single one of the 31 cases: 
$0$ balls are blue, $30$ are red, $1$ ball is blue, $29$  are red, etc, all the way through $30$ balls are blue, $0$ balls are red. Each one of these events has a probability of $\frac{1}{31}$ of occurring (assuming uniform distribution). 
So then we have $$E(X) = \sum\limits_{s \in S} \frac{1}{31} \cdot x(s) = \frac{1}{31} \sum\limits_{s \in S} x(s)$$ For each event, $x(s)$ is the probability of taking a red ball from the bag. Note that for our events, we have $40+0$ red balls, then $40+1$ red ball, $\ldots$ $40+30$ red balls. As we have $100$ balls total, this means we have $$\sum_{s \in S} x(s) = \frac{40}{100} + \frac{41}{100} + \ldots + \frac{70}{100} = \sum\limits_{i=40}^{70} \frac{i}{100} = \frac{1}{100}\sum\limits_{i=40}^{70} i$$
Then we just use the summation formula $$\sum_{i=1}^{n} = \frac{n(n+1)}{2}$$
Noting that $\sum_{i=40}^{70} i = \sum_{i=1}^{70} - \sum_{i=1}^{39}$, we then have: $$\frac{1}{100}\sum_{i=40}^{70} i = \frac{1}{100}\left(\sum_{i=1}^{70} - \sum_{i=1}^{39}\right) = \frac{1}{100}\left( \frac{70(70+1)}{2} - \frac{39(39+1)}{2}\right)$$
$$ = \frac{1}{100} (2485-780) = \frac{1705}{100} = 17.05$$
Don't forget that we're trying to find $E(X) = \frac{1}{31} \sum\limits_{s \in S} x(s)$, so multiply by $\frac{1}{31}$: $17.05 \cdot \frac{1}{31} = 0.55$
So the expected probability of taking $1$ red ball out of the bag is $0.55$.
Note that $0.55$ is the average of $0.4$ and $0.7$, the minimum and maximum probabilities for the respective cases of the minimal and maximal number of red balls in the bag. As the intermittent numbers of red balls are uniformly distributed, our result should make sense intuitively. 
A: Assuming the expected amount of indefinite red balls in the bag is 15, you have $E[r] = 40+15= 55$
and out of the total balls you have $P(r) = .55$
A: The question was edited in a substantial manner, so here's a new answer for an essentially new question:
We have a bag with $100$ balls, of which $40$ are red and $30$ are blue. The remaining $30$ "follow the distribution of the already known set". For our already known set, we have $40$ red and $30$ blue out of $70$ total balls, so the proportion of reds is $\frac47$ and the proportion of blues is $\frac37$. \
Multiplying both by $30$, we get that out of the $30$ remaining balls, $\frac{4\cdot 30}{7} = \frac{120}{7}$ are red and $\frac{3 \cdot 30}{7} = \frac{90}{7}$ are blue. So our total numbers of balls are $40+\frac{120}{7}$ red and $30+ \frac{90}{7}$ blue (which is clearly equal to $100$ balls total). 
Noting that $$\begin{align*} &40+\frac{120}{7} = 57.1429\ldots \\ &30+\frac{90}{7} = 42.8571\ldots \\ \end{align*}$$ So we are expecting $57.1429$ red balls in a bag of $100$ balls. Thus, the probability of getting a red ball out of the bag is $\frac{57.1429}{100} = 0.571429\ldots$
Finally, we should note that while this is a legitimate probabilistic result, we may find it difficult to perform inference on this result, i.e. in trying to find the expected number of red balls in the bag: then we'd actually expect  $57.1429 \approx 57 $ balls, which is approximate because I'm assuming that you can't have fractions of balls: they are integer-valued.
