91Ó°ÊÓ

There are$n$socks,$3$which are red, in a drawer.

Definition permutation ( order is important):

${}_{n}P_r=\frac{n!}{(n-r)!}$

Definition combination (order is not important):

${}_{n}C_r=\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!(n-r)!}$

with$n!=n·(n-1)·…·2·1$.

The order in which the socks are selected is not important (as a different order result in the same colors of the socks) and thus we need to use the definition of a combination. There are $\left(\begin{array}{l}n\\ 2\end{array}\right)$ways to select$2$out of$n$socks and thus there are $\left(\begin{array}{l}n\\ 2\end{array}\right)$

possible outcomes$\#\text{of possible outcomes}=\left(\begin{array}{l}n\\ 2\end{array}\right)$.

There are $\left(\begin{array}{l}3\\ 2\end{array}\right)$ways to select$2$, $3$red socks and thus there are $\left(\begin{array}{l}3\\ 2\end{array}\right)$favorable outcomes. The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$\begin{gathered} P(bothred)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}} \\ =\frac{\left(\begin{array}{l}3\\ 2\end{array}\right)}{\left(\begin{array}{l}n\\ 2\end{array}\right)} \\ =\frac{\frac{3!}{2!(3-2)!}}{\frac{n!}{2!(n-2)!}} \\ =\frac{3}{\frac{n(n-1)(n-2)(n-3)…1}{2\left(1\right)(n-2)(n-3)…\left(1\right)}} \\ =\frac{3}{\frac{n(n-1)}{2}} \\ =\frac{6}{n(n-1)} \end{gathered}$$

Next, we require the probability of both socks being red to is$\frac{1}{2}$.

$\frac{1}{2}=P\left(\text{both red}\right)$

$$\begin{gathered} \frac{1}{2}=\frac{6}{n(n-1)} \\ n(n-1)=12 \\ n(n-1)-12=0 \\ {n}^2-n-12=0 \\ (n-4)(n+3)=0 \\ n-4=0orn+3=0 \end{gathered}$$

Since the number of socks $n$in the drawers needs to be a non-negative integer, $n=-3$is impossible and thus we require that there arelocalid="1649844523030" $n=4$socks in the drawer.