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Let's say I have a small box with 2 yellow,5 green and 3 white balls. Three balls are drawn without keeping it back.How do you calculate the probabilities of the no of drawn yellow and green balls? I just want to know the logic behind this

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I'm not completely sure I understand your question. If you're only interested in the number of yellow or green balls, then you really only have two groups to worry about: The 3 white balls and the 7 green-or-yellow balls. In this case you can use the hypergeometric distribution. The probability of drawing exactly $k$ yellow-or-green balls would be

$$P(X = k) = \frac{ {{7}\choose{k}}{{3}\choose{3-k}}}{{{10}\choose{3}}}, k \in (0,1,2,3)$$

But you might be interested in the separate numbers. In that case the sampling distribution of the number of balls from each colour, drawn without replacement, is the multivariate hypergeometric distribution. The probability of drawing exactly $k$ yellow balls, $l$ green balls, and $m$ white balls, without replacement would be

$$P(X = (k,l,m)) = \frac{{{2}\choose{k}}{{5}\choose{l}}{{3}\choose{m}}}{{{10}\choose{3}}}$$

The logic for the hypergeometric distribution follows from the usual counting rules for probabilities. The multivariate hypergeometric distribution requires a bit more complicated counting, but is also straightforward.

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