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).

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).