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Suppose I am to choose six balls without replacement from a bag containing $2$ white, $2$ black and $6$ red balls. What will be the probability of at least $1$ white and $1$ black ball drawn?

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closed as off-topic by Holo, Xander Henderson, Jyrki Lahtonen, Mostafa Ayaz, Claude Leibovici Sep 14 '18 at 8:35

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    $\begingroup$ Inclusion exclusion works well here. $\endgroup$ – lulu Aug 26 '18 at 13:45
  • $\begingroup$ Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site, then edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. $\endgroup$ – N. F. Taussig Aug 26 '18 at 13:52
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The required probability is: $$\frac{{2\choose 1}{2\choose 1}{6\choose 4}+{2\choose 2}{2\choose 1}{6\choose 3}+{2\choose 1}{2\choose 2}{6\choose 3}+{2\choose 2}{2\choose 2}{6\choose 2}}{{10\choose 6}}=\frac{31}{42}.$$

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Let's calculate the number of outcomes. There are $3$ choices for the number of white balls, $0,1$ and $2$. Same is with black balls. The rest are red balls. Then, we get $3 \times 3 = 9$ outcomes.

If we have to get at least one white and one black ball, there are $2 \times 2 = 4$ choices left. Hence, the probability is $4 \over 9$.

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