## Part A

We would like to find the fewest number of socks (which are either red or black) such that the probability of two drawn socks both being red is $$\frac{1}{2}$$. Let $$r$$ be the number of red socks and $$b$$ be the number of black socks.

The probability that the two drawn socks are both red is: $\frac{r}{r+b}\cdot\frac{r-1}{r+b-1} = \frac{1}{2}$

Note that, for $$b > 0$$, $$\frac{r}{r+b} > \frac{r-1}{r+b-1}$$. Using this, we can say that:

$\left( \frac{r}{r+b} \right)^2 > \frac{1}{2} > \left( \frac{r-1}{r+b-1} \right)^2$

Taking the square root and rearranging gives:

$\frac{1-\sqrt{2}-b}{1-\sqrt{2}} > r > \frac{b}{\sqrt{2}-1}$

Trying $$b=1$$, we get $$3.414 > r > 2.414$$, which leaves only $$r=3$$. Thus, the fewest number of socks for a $$\frac{1}{2}$$ probability of drawing two reds is $$4$$ (with $$3$$ red socks, and $$1$$ black sock).

## Part B

Now, we would like to solve the same question with the added constraint that the number of black socks is even. To do so, we will try increasing even values for $$b$$ along with the inequality from Part A.

$$b$$ Inequality Potential $$r$$ Probability of two reds
2 $$4.828 < r < 5.828$$ 5 0.476
4 $$9.656 < r < 10.656$$ 10 0.494
6 $$14.485 < r < 15.485$$ 15 0.50

Thus, The fewest number of socks with an even number of black socks is 21 (with 15 reds and 6 blacks).