# The Sock Drawer: Probability and Statistics Problem

#### b) How small if the number of black socks is even?

SOLUTION:

Let $r$ red and $b$ black socks. The probability of the first sock’s being red is $\frac{r}{r+b}$; and if the first sock is red, the probability of the second’s being red now that a red has been removed is $\frac{r-1}{r+b-1}$. Then we required the probability that both are red to be $\frac{1}{2}$, or $\frac{r}{r+b}\times \frac{\:r-1}{r+b-1}=\frac{1}{2}$

Notice that $\frac{r}{r+b}>\frac{\:r-1}{r+b-1}$

Therefore, we can create the inequalities $\left(\frac{r}{r+b}\right)^2>\frac{1}{2}>\left(\frac{\:r-1}{r+b-1}\right)^2$

Taking the square roots, we have, for r>1. $\frac{r}{r+b}>\frac{1}{\sqrt{2}}>\frac{\:r-1}{r+b-1}$

Simplifying, we have $\left(\sqrt{2}+1\right)b+1>r>\left(\sqrt{2}+1\right)b$

So we can now easily plug in values for b, then solve for r.

When $b=1$, r must be between 2.414 and 3.414, and so $r=3$. For $r=3,\:b=1$, $P\left(2\:red\:socks\right)=\frac{3}{4}\cdot \frac{2}{3}=\frac{1}{2}$ $Therefore,\:the\:smallest\:number\:of\:socks\:is\:4.$.

Using the same inequality, we can substitute even values for b starting from 2, then solve for the value of r. After, check if the probability is 1/2. Refer to the table below.

So, when b is 6, r is 15 and the probability of getting 2 red socks is 1/2. This condition satisfies the problem. $Therefore,\:21\:is\:the\:smallest\:number\:when\:b\:is\:even.$.