Probability of Drawing Red Socks
A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is \( \frac{1}{2} \). (a) How small can the number of socks in the drawer be? (b) How small if the number of black socks is even?
Let us begin by making the reasonable assumption that each draw is independent. Then let
- \( R \) : denotes the event of drawing a red sock
- \( r \) and \( b \) denote the number of red socks and black socks respectively
Then it follows that
$$ \mathbf{Pr}(R,R) = \frac{r}{r+b}\frac{r-1}{r+b-1} $$
Then it is easy to see that if you have four socks in total with 3 red and 1 black, then \( \mathbf{P}(r,r) = \frac{3}{4}\frac{2}{3} = \frac{1}{2} \). Furthermore, note that if there were 2 red and 2 black then \( \mathbf{Pr}(R,R) = \frac{2}{4}\frac{1}{3} \neq \frac{1}{2} \). We claim this is the minimal case, to see this note that if there is 1 red and 3 black then obviously the drawing two red socks is a probability 0 event. If we have 3 socks in total, then we must have 2 red socks and 1 black sock, otherwise, it is clear that we have either a probability 1 event or a probability 0 event, however in this case note that \( \mathbf{P}(r,r) = \frac{1}{3} \). Furthermore, it is also clear that one cannot have less than 3 socks in total, as if there were only one sock in total then \( \mathbf{P}(r,r) = 0 \), and if there were two socks in total either \( \mathbf{P}(r,r) = 0 \) or \( \mathbf{P}(r,r) = 1 \). Hence the correct answer for part (a) is \( 4 \).
The answer to part (b) is slightly more complicated. First note that by the above, the pair \( (r,b) \) must satisfy the equation:
$$ \frac{1}{2} = \frac{r}{r+b}\frac{r-1}{r+b-1} \iff r^2 - r(1+2b) + (b-b^2) = 0 $$
using the quadratic formula:
$$ r = \frac{1}{2} + b \pm \frac{1}{2}\sqrt{1 + 8b^2} \tag{1} $$
and substituting \( b = 2x \) as we demand \( b \) to be even we obtain:
$$ r = \frac{1}{2}\left[ (1+4x) \pm \sqrt{1 + 32x^2} \right] \tag{2} $$
we note that \( r \) must be an integer, hence we must have that \( (1+4x) \pm \sqrt{1 + 32x^2} \) is an even integer. The smallest \( x \) to accomplish this is \( x = 3 \) by easy case work, when taking the \( + \) version. Furthermore, it is easy to justify that the correct solution to take is the \( + \) version. To see this recall from the above that \( r = 3 \) and \( b = 1 \) is a valid solution, hence these values should satisfy (1), then \( 3 = 1.5 \pm 1.5 \), hence clearly one must take the adding solution. Therefore, we have that the smallest even number of black socks we can have is \( b = 2x = 6 \) and by (2) this determines the number of red socks we can have, namely \( r = 15 \), thus the smallest total number of socks in the box is \( 21 \).