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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?

Rods2292
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    Actually, $6\cdot (\frac{2}{n}\cdot \frac{1}{n-1}\cdot \frac{n-2}{n-2}\cdot \frac{n-3}{n-3})$ is the probability that the two white balls are in the sample. Meanwhile the middle expression you calculated is the probability that no white ball is in the sample. You did not however use the sentence "... 2 white balls are in the sample is 6 times higher than...no white balls ...". To fix this, the middle expression should be multiplied by $6$. You have also an "$=8$" which does not belong. The far left expression is a probability and cannot equal eight. Use equals only for equality. – JMoravitz May 11 '16 at 00:29
  • I edited it. It was supposed to be n = 8. – Rods2292 May 11 '16 at 22:44

2 Answers2

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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 $n=6$.

Brian M. Scott
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  • "your lefthandside is too small by a factor of 6." Perhaps the six written is in fact the $\binom{4}{2}$ term which accounts for the fact that the two white balls need not be picked first. Regardless, is it not the middle expression which is too small by a factor of six since it is the middle expression relating to the probability of no white balls which should be magnified to compare to the probability of both white balls? – JMoravitz May 11 '16 at 00:34
  • @JMoravitz: No, I simply reversed the sides (and forgot that there was an unexplained $8$ on the end. I'm typing this on a Kindle and have a very limited field of view. – Brian M. Scott May 11 '16 at 00:45
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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.

Fimpellizzeri
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