Q. 2.38 - A First Course in Probability

Question

There are$n$ socks, $3$which are red, in the drawer. What is the value of n if, when $2$the socks are chosen randomly, the probability that they are both red is$\frac{1}{2}$?

Step-by-Step Solution

Answer

The probability that they are both red$\frac{1}{2}$ is$n=4$.

Step 1 Given Information.

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

Step 2 Explanation.

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$n!=n·(n-1)·\dots ·2·1$.

Step 3 Explanation.

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:

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

Step 4 Explanation.

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

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

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

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 are$n=4$ socks in the drawer.