Confidence interval and sample size multinomial probabilities I'm an absolute beginner in statistics. Please excuse any wrong assumptions or missing information in my question.
I have a question that relates to a multinomial distribution (not even 100% sure about this) that I hope somebody can help me with.
If I take a sample (lets assume $n=400$) on a categorical variable that has more than two possible outcomes (e.g. blue, black, green, yellow) and plot the frequencies so that I can derive the probabilities. E.g.:
black 10%
blue 25%
green 35%
yellow 30%
How could I compute the 95% confidence interval for those probabilities? 
And how could I determine the sample size required in order to get an accurate result within +-3% for each probability?
Please let me know if the answer to the question requires any additional information.
 A: Thank you very much again for your help. Below is the (hopefully correct) solution using the "Normal Approximation Method" of the Binomial Confidence Interval:

A: I would like to add Wilson's method mentioned by Michael M in a comment.
From Wikipedia: Binomial proportion confidence interval - Wilson_score_interval.
You can get a 95% confidence interval by using the following: 
$\frac{n_s + \frac{z^2}{2}}{n+z^2} \pm \frac{z}{n+z^2}\sqrt{\frac{n_s n_f}{n}+\frac{z^2}{4}}$ 
The left term is the center value and the right term gives the value you have to add / subtract to get the interval bounds.
$n_s$ is the number of samples in that category, $n_f$ the number of samples not in that category, $n$ the total number of samples and $z$ is 1.96 if you want a 95% confidence interval*
For high counts it gives the same results of the normal approximation, however this should be better for low counts or extreme values.  
As an example, I had a category with 0 samples and the normal approximation returned a 0 s.e., and thus an absured confidence interval of 0-0 (as it was certain that it has 0% probability, while actually it had 0 occurence only because of the few samples).
$ $ 
* The method is actually for a binomial distribution and $n_s$ and $n_f$ are successes and failures on that distribution.
However, I think it can be reasonably used for a multinomial, even though it does not account for the fact that estimated probabilities must sum up to 1.
The normal approximation doesn't account for that either afaik.
