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Suppose a basket contains $B$ blue balls and $R$ red balls. Then, suppose I pick $N$ balls from the basket. What is the probability that I get

  1. At least $b$ blue ones and $r$ red ones
  2. At least $b$ blue ones and exactly $r$ red ones
  3. Exactly $b$ blue ones and exactly $r$ red ones

How does the solution differ if I put the balls back or don't?

I suppose this is a very basic problem in statistics so I attempted to cover a range of related problems here. I suppose I should know about permutations and combinations for solving this. (This is not a homework, I'm studying stats for fun.) What I am trying to learn here is how to solve any similar kind of problem.

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closed as too broad by whuber Jul 25 '16 at 15:30

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ The easiest way to learn this kind of stuff is to choose small numbers e.g. $B=2, R=3, N=4$ or maybe $N=6$ and make a list of all possible outcomes for small values of $b$ and $r$. That will help you understand what things need to be considered in finding the answer in general. $\endgroup$ – Dilip Sarwate Mar 23 '15 at 14:46
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    $\begingroup$ The keywords are "binomial distribution" and "hypergeometric distribution". $\endgroup$ – Elvis Mar 23 '15 at 15:05
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If you are looking to get "at least" b balls then I assume you want a 100% probability which means N should be equal to all r balls plus the number of b balls you've set as the minimum. This guarantees the minimum for b and you will also have exactly (N - minimum b) x r / (all balls).

When you draw a ball you statically get a fraction of each color, so N = (all b) /(total) + (all r) / (total).

It doesn't matter if you put the balls back from a statistical point of view - you will always have the same odds if you always have the same ball group. When you don't put the balls back then you have the chance of changing the probabilities but they still work by the same formula.

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    $\begingroup$ I believe you may have misinterpreted the question. It is asking a probability problem rather than a statistical problem. For instance, with $N=1$, $B=50$, and $R=50$, the chance of getting at least $b=1$ blue balls and $r=0$ red balls is $50\%$. The correct answer will refer to binomial and hypergeometric distributions, as @Elvis indicated in a comment. $\endgroup$ – whuber Mar 24 '15 at 14:44
  • $\begingroup$ Thank you, I stand corrected. I did misinterpret the question. $\endgroup$ – NuCoder Mar 24 '15 at 15:59

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