# probability of choosing the same color twice in 5 tries

I'm sure this question has been asked before but I couldn't find an answer.

There are 8 balls: 3 red, 2 blue and 3 black. Whats the probability of choosing at least 2 red balls if we pick 5 balls?

I would like to know the answer and how to solve those problems.

Thank you.

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Think about the event. This can be done in two ways basically. Choosing 2 red or 3 red.

2 red: You can choose these in $\binom{3}{2}=3$ ways. You are left with 3 balls to choose from the other 5, there are $\binom{5}{3}=10$ ways to do this. So 30 ways in total.

3 red: You have to pick all red ones (there is only one way to do it). $\binom{5}{2}= 10$ ways to choose the rest.

Ok, now you need to count all possible ways of choosing 5 balls from a pool of 8: this is $\binom{8}{5}=56$.

So the answer is $\frac{10+30}{56}=\frac{40}{56} = 0.71 \cdots$.

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are you sure the answer is correct? You said "you are left with 3 balls to choose from the other 5". Aren't there 6 balls left instead of 5? Also, is there only one way to pick all 3 red balls? the first 3 balls that you pick can be red or the last 3 balls can be. Aren't there more then one way? Please expand the answer a little bit. – Ionut Hulub Nov 25 '12 at 5:01
yeah, but the 6th one is red and you don't want to choose it cause otherwise you would be choosing 3 red instead of 2 and you are examining the "2 red" case. the answer is correct... – chango Nov 26 '12 at 9:18
of course there is only one way to pick 3 red balls out of a pool of 3 red balls, you need to pick them all! if not compute $\binom{3}{3}=1$. – chango Nov 26 '12 at 9:34