Determining the number of balls in a box I'm facing problems solving this question and I'd like some help: 
A box contains n balls, where just 2 are white and the rest are red. A random sample of 4 balls is drawn without replacement. It's known that the probability of the 2 white balls are in the sample is 6 times higher than the probability that no white balls are in the sample. Calculate n.
I did like that:
$$6* (\frac{2}{n} *\frac{1}{n-1} * \frac{n-2}{n-2} * \frac{n-3}{n-3}) = (\frac{n-2}{n} *\frac{n-3}{n-1} * \frac{n-4}{n-2} * \frac{n-5}{n-3}) => n = 8 $$
But, according to the answer of this question, n = 6.
I trying to find my mistake. Can someone help me?
 A: Your mistake is that you are assuming that the two white balls are picked first, when in fact there are $\binom{4}2=6$ pairs of positions in which they could be chosen. Thus, your lefthand side is too large by a factor of $6$. (I'm ignoring the $8$ on the end, since it is clearly impossible.)
I think that it's a bit easier to count the outcomes using combinations. There are $\binom{n-2}2$ samples that contain both white balls, and $\binom{n-2}4$ that contain no white ball. The probability of getting a sample of either kind is proportional to the number of possible samples of that kind, so $\binom{n-2}2=6\binom{n-2}4$. Expanding this yields
$$\frac{(n-2)(n-3)}2=\frac{(n-2)(n-3)(n-4)(n-5)}4$$
and then $2=(n-4)(n-5)$. The positive solution to this quadratic is $n=6$.
A: They wat you're doing it, the order in which you draw the balls makes a difference (first you draw two white balls, then two red balls).
Rather, think of it as follows:


*

*How many possible combinations of four balls are there? That would be $\binom{n}{4}$.

*How many possible combinations of four balls, two of which are white, are there? Well, you must take the only $2$ white balls and also $2$ red balls from the remaining $n-2$ balls, for a total of $\binom{n-2}{2}$.

*How many possible combinations of four balls, all of which are red, are there? You must take $4$ balls from the $n-2$ red balls, for a total of $\binom{n-2}{4}$.
Hence, the probability of drawing two white balls and two red balls is
$$\frac{\binom{n-2}{2}}{\binom{n}{4}}=\frac{12}{n(n-1)}$$
And the probability of drawing four red balls is
$$\frac{\binom{n-2}{4}}{\binom{n}{4}}=\frac{(n-4)(n-5)}{n(n-1)}$$
So that $12=6(n-4)(n-5)$. You can check that $6$ is the only positive integer solution.
