Question 320792
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Let *[tex \Large x] represent the number of blue socks in the drawer before the first selection is made.  Then we know the probability of selecting a blue sock on the first trial is *[tex \Large \frac{x}{6}].  Given that a blue sock is selected on the first trial, we know that there are one fewer total socks and one fewer blue socks, hence the probability of drawing a blue sock on the second trial is: *[tex \Large \frac{x\ -\ 1}{5}].  Then the overall probability of drawing two blue socks is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{x}{6}\right)\left(\frac{x\,-\,1}{5}\right)\ =\ \frac{2}{3}]


A little algebra gets us to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2\ -\ x}{30}\ =\ \frac{20}{30}]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 20\ =\ 0]


Which factors to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 4)(x\ -\ 5)\ =\ 0]


Toss out the negative root (minus 4 blue socks?  Ludicrous!) and we see that we must have 5 blue socks out of 6 at the start in order to be only 2/3 certain of drawing a pair.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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