# Probability of drawing a pair of brown socks

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

• 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. – Jubbles May 27 '15 at 20:42
• is this a homework problem? – Jubbles May 27 '15 at 20:45

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.

• That is the best approach to this question – jumetaj May 27 '15 at 18:06
• +1 - yes, this is a great answer. Curious, shouldn't the formating be $$\frac{10 \choose 2}{16 \choose 2}$$ ? – JTP - Apologise to Monica May 28 '15 at 0:50
• @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. – Docteur May 28 '15 at 9:03
• The notation is fine. – VividD May 28 '15 at 9:05
• @Docteur - thx, the wikipedia article shows it and a few others, all being equivalent. – JTP - Apologise to Monica May 28 '15 at 9:52

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

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

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