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What is the probability of selecting all three colors when $4$ balls are randomly selected from a box containing $7$ black, $4$ green and $2$ yellow balls?

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    $\begingroup$ Hello @xxssxxdd, welcome to MSE. Questions in the style "here is the question, solve it" are not always well-received on MSE. Can you please add in the question what your attempts are? $\endgroup$ – Ernie060 Nov 10 '18 at 10:49
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you can use combination formula to find out:$$p(3C)=p(2B+1G+1Y)+p(1B+2G+1Y)+(1B+1G+2Y)$$ $$=\frac{\binom{7}{2}\times \binom{4}{1}\times \binom{2}{1}}{\binom{13}{4}}+\frac{\binom{7}{1}\times \binom{4}{2}\times \binom{2}{1}}{\binom{13}{4}}+\frac{\binom{7}{1}\times \binom{4}{1}\times \binom{2}{2}}{\binom{13}{4}}$$ $$=\frac{56}{143}$$ Where $3C=$three colors, $B$=number of black ball,$G=$number of green ball,$Y$=number of yellow ball

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You need to specify if you pick the three balls at the same time, or if you put them back before you pick the next one.

Edit : Sorry, the information was present but had been removed by the first reviewer... :-/

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