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} $ ?
Best Answer
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. $1-2\cdot \frac{\binom{n}{2}}{\binom{2n}{2}}=\frac{n}{2n-1}$