# [SOLVED]Finding n number of socks given a probability

#### Jameson

Staff member
I have checked and rechecked my thought process, but can't seem to figure this one out.

There are n socks, 3 of which are red. What is the value of n if 2 of the socks are chosen at random, the probability that both are red is 0.5?

This seems like a fairly straightforward combination problem. Here's my work.

$$\displaystyle \left( \frac{\binom{3}{2}}{\binom{n}{2}}=\frac{1}{2} \right) \rightarrow \left( \frac{3}{\frac{n(n-1)}{2!}}=\frac{1}{2} \right) \rightarrow n(n-1)=3$$

Solving that quadratic leads to non-integer solutions so we have a problem. Where's my error?

EDIT: Even approaching it another way I get the same thing. Let's not use combinations. Instead say the probability of choosing 1 red sock is $$\displaystyle \frac{3}{n}$$ and after that the probability of choosing another red sock is $$\displaystyle \frac{2}{n-1}$$. We get the same thing $$\displaystyle \frac{3}{n} \frac{2}{n-1} = \frac{1}{2}$$

#### MarkFL

Staff member

$$\displaystyle \left( \frac{\binom{3}{2}}{\binom{n}{2}}=\frac{1}{2} \right) \rightarrow \left( \frac{3}{\frac{n(n-1)}{2!}}=\frac{1}{2} \right) \rightarrow n(n-1)=3$$

$n(n-1)=12$

which has integral roots.

#### Jameson

Staff member
Hmm, I appreciate the correction but I still can't spot the error in my work. If the math from my OP you quoted has 3 expressions, in which expression did I first make a mistake. I just can't see it .

#### MarkFL

Staff member
We have:

$\displaystyle\frac{3}{\frac{n(n-1)}{2}}=\frac{1}{2}$

Next, on the left, we may bring the 2 in the denominator up top as follows:

$\displaystyle\frac{6}{n(n-1)}=\frac{1}{2}$

Cross-multiply:

$n(n-1)=12$

You were essentially trying to use:

$\displaystyle \frac{1}{\frac{1}{2}}=\frac{1}{2}$

and this led to the RHS of you quadratic being 1/4 what it should be.

#### Jameson

This kind of small thing can be maddening! Thanks again. Very glad to offer a second pair of eyes! 