For a Given a Probability of Success, How Many Successes in a Sample?

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The discussion revolves around calculating the number of blue balls needed in a sack of 100 to achieve at least 3 blue balls with an 85% success rate when drawing 10 balls. The initial calculations suggested needing about 43 blue balls, but further analysis indicated that 40 blue balls would provide approximately an 84.6% chance of success. The problem is framed as an inverse hypergeometric distribution challenge, with participants seeking a general equation to solve for the number of successes needed. The conversation emphasizes the need to correctly apply hypergeometric distribution principles rather than binomial distribution. Ultimately, the conclusion is that having 40 blue balls meets the desired probability threshold for the scenario described.
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Is there a general equation for an inverse hypergeometric distribution?

Greetings,

I'm making a statistical calculator for game analysis, and have an interesting problem.

Here is a specific example: I have a sack of 100 balls and are going to take 10 out of it randomly. I want to find 3 or more blue balls 85% of the time.

How many blue balls must I have in the sack to have this probability of success?
X = 100
Y = ?
K = 10
R = 85%
N = 3

In general terms:
To achieve at least N successes R percent of the time in a sample size of K, how many items Y in a set of X items must be successes?

I arrived at a general equation by my own calculations to find N given R, but it indicates I need about 43 blue balls get achieve my desired 85% chance.

However, using a standard cumulative hypergeometric distribution to find R given N, I calculate that using 43 balls will give me a 88.9% chance of success.

Consequently, I know my method is in error, and am hoping I can have some help in figuring a general equation for this problem. It would seem what I am seeking is an inverse hypergemoetric distribution.

The equations and work so far can be seen in the google docs spreadsheet I have created for them http://spreadsheets.google.com/ccc?key=0AnPw5qvi2hRrdHoweklSQnBuVW9NbVFIUENpYmUyV3c&hl=en".
You can see my work on the Chance to Draw Stats and Probability tabs.

Any input would be appreciated.
 
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Having 40 blue balls gives you a 0.846ish chance of having 3 or more blue balls. In general the problem reduces to solving a polynomial equation. Do you know how to set up the equation?
 
Classic binomial distribution.
 
zli034 said:
Classic binomial distribution.

I was wrong. It's a hypergeometric distribution problem. I have calculated if you want 84.6% probability to get 3 or more blue balls from random picked 10 balls out of 100 balls, the 100 balls need to have at least 40 blue balls and also 60 other colored balls.
 
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