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

  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ is this a homework problem? $\endgroup$
    – Jubbles
    May 27, 2015 at 20:45

3 Answers 3


The easier solution (for understanding) would be:


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.

  • $\begingroup$ 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

$$\dfrac {10}{16} \cdot \dfrac{9}{15} = .375$$

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.