# Finding the probability that at least $1$ white and $1$ black are drawn when six balls are selected from a bag [closed]

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?

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

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## 2 Answers

The required probability is: $$\frac{{2\choose 1}{2\choose 1}{6\choose 4}+{2\choose 2}{2\choose 1}{6\choose 3}+{2\choose 1}{2\choose 2}{6\choose 3}+{2\choose 2}{2\choose 2}{6\choose 2}}{{10\choose 6}}=\frac{31}{42}.$$

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 $4 \over 9$.