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Suppose there are 5 black, 10 white, and 15 red marbles in an urn for a total of $N=5+10+15$ marbles. You randomly draw $n=6$ marbles without replacement. The probability that you pick exactly two of each color is given by the multivariate hypergeometric function.

$$\Pr[1\textrm{ black},1\textrm{ white},2\textrm{ red}]=\frac{\binom{5}{2}\binom{10}{1}\binom{15}{1}}{\binom{30}{6}} $$

How is possible instead in Mathematica to get the probability to draw at least 1 black, 1 white and 2 red?

$$\Pr[\textrm{ black} \geq 2, \textrm{ white} \geq 2, \textrm{ red}\geq 2] $$

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    $\begingroup$ If you draw 6 marbles and you want the probability of at least two of each, this can only happen by having exactly two of each. $\endgroup$
    – bill s
    Commented May 4, 2016 at 13:39
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    $\begingroup$ An urn contains 12 red balls, 23 blue balls, and 9 green balls see on Applications $\endgroup$
    – user9660
    Commented May 4, 2016 at 13:54
  • $\begingroup$ I came up with this solution: [ScriptCapitalD] = MultivariateHypergeometricDistribution[5, {12, 23, 9}]; Probability[ x > 1 [And] y > 1 [And] z > 0, {x, y, z} [Distributed] [ScriptCapitalD]] $\endgroup$
    – linello
    Commented May 4, 2016 at 14:41

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