Is is possible to estimate the distribution by drawing observations? I want to estimate how many red balls in a box. Red, yellow, blue balls could be in the box. But I don't know how many of them are in the box.
What I did was randomly drawing 10 balls from the box and learned that there was no red ball.
(Edit: Assume the number of the balls in the box is a known finite number. Let's say 10,000)
Can I say the possibility of having at least one red ball is $$\left(\frac13\right)^{10}= 0.00169\ \%\ ?$$ (3 possible outcomes, and 10 observations)
I'm thinking I don't know the distribution of the colors of the balls, so I'm not sure this inference is reasonable or not. Thanks!
 A: No, you cant say that. 

Infinite case: 

The reason is that you dont know how many balls there are in a box in total.  For instance, it may be that a box contains infinite number of balls (since we are talking about mathematical box, I can make this assumption) $\frac{9}{10}$'th of which are red.  In this case, the probability of drawing at least $1$ red ball out of $10$ is will be different from what you calculated.

Finite case:

Similar argument applies.  If it happens that the true distribution of balls are such that the share of red balls is $9/10$, then your above calculation is not correct.
A: TL;DR: The answer to the title of your question is, "Yes, but not with anywhere near the kind of accuracy and confidence you seem to be asking for."

Let's say I put $10000$ balls in a box, and exactly one of the balls is red.
You know the total number of balls but not how many there are of each color.
You draw $5000$ balls without replacement and look at each one to see if it is red.
I then ask you whether there was a red ball in the box before you started taking balls out, and you answer "yes" or "no."
If your criterion for "no red ball" is that none of the balls you drew were red, then there is a $50\%$ chance you will give me the wrong answer.
Now consider the case where you draw only $10$ balls. If you think getting only yellow and blue balls in the first ten is enough to say that there is not at least one red ball in the box, what's the probability that you'll give the wrong answer in the case where there actually is exactly one red ball in the box?
Note that it doesn't matter how many other colors there are. You'll get wrong answers just as likely when we fill the rest of the box with balls of $100$ different colors as when we use only yellow and blue balls in addition to the red ball.

Trying to decide there are no individuals of type X in a large population is something you basically cannot resolve simply by examining a small sample.
The formula you give would be a reasonable one to use if the question you are trying to answer is, "Are at least $\frac23$ of the balls red?"
(And even then, I would not describe your result as the probability that the answer is "yes".)
