MHB Finding n number of socks given a probability

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The discussion revolves around solving a probability problem involving socks, specifically determining the total number of socks (n) given that the probability of randomly selecting two red socks from three is 0.5. The initial approach using combinations led to the incorrect equation n(n-1)=3, resulting in non-integer solutions. A correction pointed out that the proper formulation should yield n(n-1)=12, which has integer roots. The error stemmed from an incorrect cancellation during the calculations, highlighting the importance of careful mathematical manipulation. The participants express appreciation for collaborative problem-solving in mathematics.
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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.

[math]\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[/math]

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 [math]\frac{3}{n}[/math] and after that the probability of choosing another red sock is [math]\frac{2}{n-1}[/math]. We get the same thing [math]\frac{3}{n} \frac{2}{n-1} = \frac{1}{2}[/math]
 
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Your problem is here:

[math]\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[/math]

This instead leads to the quadratic:

$n(n-1)=12$

which has integral roots.
 
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 (Headbang).
 
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.
 
I got it now. Thank you MarkFL! I was doing some kind of incorrect cancellation with the two in a numerator and denominator across the equals sign instead of cross-multiplying.

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