Conditional probabilities, urns I found this question interesting and apparently it has to do with conditional probabilities:
An urn contains six black balls and some white ones. Two balls are drawn simutaneously. They have the same color with probability 0.5. How many with balls are in
the urn?
As far as I am concerned I would say it is two white balls...
 A: probability of 2 black balls: $\frac{6}{8}*\frac{5}{7}=\frac{30}{56}$
of 2 white balls: $\frac{2}{8}*\frac{1}{7}=\frac{2}{56}$
$\frac{30}{56}+\frac{2}{56}=\frac{32}{56}\neq\frac{1}{2}$
Assume the number of white balls is $n$:
2 black balls = $\frac{6}{6+n}*\frac{5}{5+n}=\frac{30}{n^2+11n+30}$
2 white balls = $\frac{n}{6+n}*\frac{n-1}{5+n}=\frac{n^2-n}{n^2+11n+30}$
total probability $=\frac{1}{2}$ (as per problem setup) and also $=\frac{30}{n^2+11n+30}+\frac{n^2-n}{n^2+11n+30}=\frac{n^2-n+30}{n^2+11n+30}$
Moving sides, $2(n^2-n+30)=n^2+11n+30$
$2n^2-2n+60=n^2+11n+30$
$n^2-13n+30=0$
$(n-10)(n-3)=0$
$n=\{10,3\}$
Thus, this problem is solved when there are either 3 or 10 white balls.
A: Let there be $6$ black and $w$ white balls. The probability $P$ that we draw a pair of equally colored balls is given by
$$P={{6\choose 2}+{w\choose2}\over{6+w\choose2}}={6\cdot 5+w(w-1)\over(6+w)(5+w)}\ .$$
The condition $P={1\over2}$ leads to the quadratic equation $w^2-13w+30=0$ with the two solutions $w=3$ and $w=10$.
A: Long hint/walkthrough: Let the number of white balls be denoted $w$. The probability of pulling two white balls will be $\frac{w}{6+w}\cdot\frac{w-1}{6+w-1}$ since the probability of choosing a white ball will be $P(w_1)=\frac{w}{w+6}$ and since there is one less white ball the probability of choosing another will be $P(w_2)=\frac{w-1}{6+w-1}$. To find the probability that both these events will occur we multiply their probability $P(w_1\cap w_2)=P(w_1)\cdot P(w_2)$. Note that this is only the probability of finding two white balls. What will be the probability of finding two black balls? To find the probability that one OR another event occurs we add the probability of each event ($P(x\cup y)=P(x)+P(y)$). So what is the probability of choosing two of the same color balls?
Can we find a mathematical way to express the probability of choosing one ball of each color? If so, we can set these equations equal and have $P(w_1\cap w_2)+P(b_1\cap b_2)=P(b_1\cap w_2)+P(w_1\cap b_2)$. Since our only variable should be $w$ this will allow us to solve.
