# An urn contains $n$ red and $n$ blue balls. If two balls are drawn at random from the urn, what is the probability they have different colors?

An urn contains n red and n blue balls. Two balls are drawn from the urn together, at random. What is the probability that two balls of different colors are drawn?

My approach:

The space room is:

$$\Omega$$ ={(r,b),(b,b),(r,r)} with r="red ball" and b="blue ball"

How do I calculate now the probability I'm looking for?

Is it:

$$\mathbb{P}((r,b))$$ = $$\frac{1}{2n}*\frac{1}{2n}=\frac{1}{4n^2}$$ ?

How many balls are there? $$2n$$ right? . So in how many ways can you possibly choose two balls from it? . You can do it by $$\binom{2n}{2}$$. But if you want the two balls two be different. Then you choose one ball out of $$n$$ red ball by $$\binom{n}{1}$$ and another one simultaneously from $$n$$ blue balls by $$\binom{n}{1}$$ . So the probability is $$\frac{\binom{n}{1}\cdot\binom{n}{1}}{\binom{2n}{2}}=\frac{n}{2n-1}$$.
Alternatively you can try to substract from $$1$$ , the probability of getting both balls of same colour to get the same answer. i.e. $$1-2\cdot \frac{\binom{n}{2}}{\binom{2n}{2}}=\frac{n}{2n-1}$$