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[SOLVED] Finding n number of socks given a probability

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Jameson

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Jan 26, 2012
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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.

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

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

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Feb 24, 2012
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Your problem is here:

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

This instead leads to the quadratic:

$n(n-1)=12$

which has integral roots.
 
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Jameson

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Jan 26, 2012
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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).
 

MarkFL

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Feb 24, 2012
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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.
 
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Jameson

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Jan 26, 2012
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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. :)
 

MarkFL

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Feb 24, 2012
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Very glad to offer a second pair of eyes! (Smile)