A drawer contains a number of red and blue socks.  If I pull two
out at random, then the chance of them being a red pair is a half
and the chance of them being a blue pair is a twelfth.  How many
socks are in the drawer?
 
 

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                  THE SOLUTION
 

There are 120 socks in the drawer, 85 red ones and 35 blue ones.

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My solution follows:

For my first try, letting R be the number of red socks,
and B be the number of blue socks, yielded the following
two equations:

  R    (   R - 1   ) = 1   and     B    (   B - 1   ) =  1
R + B     R + B -1     2         R + B     R + B -1     12

Simplifying you get

R ( R - 1 ) = 6B ( B - 1 )

Unless you test possible values for B starting with 2, 3, 4, etc.,
I do not know how to finish the problem.

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For my second try, I concluded that if red then red is 1/2 and
if blue then blue is 1/12 then red then blue is 5/24.  This
yielded the following equation:

  R    (     B     ) =  5
R + B     R + B -1     24

combining this third equation with either of the first two
equations again produced the same resulting equation of

R ( R - 1 ) = 6B ( B - 1 )

I will go back to the drawing board for another plan.

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For my third try at 3:00 A.M, using the derived formula of
     R ( R - 1 ) = 6B ( B - 1 ),
I am reverting to my first try, and substitution for B.

Letting B = 2 yielded R = 4 and the total of 6.
     The probability of two red socks is 4/6 * 3/5 which is 40%.

Letting B = 3 did not yield a solution.

Letting B = 4 yielded R = 9 and the total of 13.
     The probability of two red socks is 9/13 * 8/12 which is 46.2%.

Noting that we are getting closer to 50%, we marched on.

Letting B = 5 to 14 yielded no solutions.

Letting B = 15 yielded R = 36 and the total of 51.
     The probability of two red socks is 36/51 * 35/50 which is 49.4%.

Noting that we are getting closer to 50%, we continued to march on.

Letting B = 16 to 34 yielded no solutions.

Letting B = 35 yielded R = 85 and the total of 120.
     The probability of two red socks is 85/120 * 84/119 which is 50%.

The problem with this technique is that you need a calculator to
solve it.  This offends my math conscience.