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You have a drawer with $6$ loose blue socks, and $10$ loose brown socks. If you grab two socks from the drawer in the dark (random draw), what is the probability that you draw a brown pair?

I have $\frac{5}{8}=\frac{10}{16}=.625$.

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    $\begingroup$ Within the confines of the comments instructions - "suggest improvements" - the title of your question 'probability and math' could be improved. I viewed this page with the hopes of some grand question (and resulting discussion) about probability and how math provides structure around the concept, and instead found a high-school combinatorics problem. $\endgroup$
    – Jubbles
    May 27, 2015 at 20:42
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    $\begingroup$ is this a homework problem? $\endgroup$
    – Jubbles
    May 27, 2015 at 20:45

3 Answers 3

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The easier solution (for understanding) would be:

$$P=\frac{C_2^{10}}{C_2^{16}}=\frac{\frac{10!}{2!(10-2)!}}{\frac{16!}{2!(16-2)!}}=0.375$$

That is, what you want is to pick 2 of those 10 brown socks, and it is possible that you choose any 2 of all 16 socks.

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  • $\begingroup$ That is the best approach to this question $\endgroup$
    – jumetaj
    May 27, 2015 at 18:06
  • $\begingroup$ +1 - yes, this is a great answer. Curious, shouldn't the formating be $$\frac{10 \choose 2}{16 \choose 2}$$ ? $\endgroup$ May 28, 2015 at 0:50
  • $\begingroup$ @JoeTaxpayer According to Wikipedia and my memory of math classes, this is the notation used in some texts---I can confirm that it is standard in countries such as France. $\endgroup$
    – Docteur
    May 28, 2015 at 9:03
  • $\begingroup$ The notation is fine. $\endgroup$
    – VividD
    May 28, 2015 at 9:05
  • $\begingroup$ @Docteur - thx, the wikipedia article shows it and a few others, all being equivalent. $\endgroup$ May 28, 2015 at 9:52
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$$\dfrac {10}{16} \cdot \dfrac{9}{15} = .375$$

$\dfrac {10}{16}$ for first sock brown, then $\dfrac 9{15}$ for second.

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$$P(\text{First Brown }\cap \text{Second Brown}) = P(\text{First brown})\cdot P(\text{Second Brown}\ |\ \text{First Brown})$$ $$= \frac{10}{16}\cdot \frac{9}{15}$$

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