Finding n number of socks given a probability

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Discussion Overview

The discussion revolves around a probability problem involving socks, specifically determining the total number of socks (n) given that 3 are red and the probability of randomly selecting 2 red socks is 0.5. The focus is on mathematical reasoning and problem-solving techniques.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents a combination approach to find n, leading to the equation \( n(n-1) = 3 \), but finds non-integer solutions.
  • Another participant corrects the first participant's equation to \( n(n-1) = 12 \), suggesting this has integral roots.
  • A participant expresses confusion about where the initial error occurred in their calculations.
  • Further clarification is provided on manipulating the equation, indicating a mistake in cancellation rather than proper cross-multiplication.
  • A later reply acknowledges the correction and expresses gratitude for the assistance in resolving the misunderstanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial error in the calculations, but there is agreement on the corrected equation \( n(n-1) = 12 \) leading to integral solutions.

Contextual Notes

The discussion highlights the importance of careful manipulation of equations in probability problems, with specific attention to the steps taken in deriving the quadratic equation.

Jameson
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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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