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Re: There are four pairs of shoes in a big black sack: blue, [#permalink]
mciatto wrote:
1/7.

The first pick does not matter, and for the second pick there is a 1/7 chance you get the matching shoe to the first!


P=1*1/7=1/7. A nice, concise, and correct approach. A hypergeometric forfuma is also legal here.

for example, for blue shoes P= 2C2/8C2
for a pair of any color P=4*2C2/8C2=1/7
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Re: There are four pairs of shoes in a big black sack: blue, [#permalink]
I agree with mciatoo. An excellent an consise solution. :)
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Re: There are four pairs of shoes in a big black sack: blue, [#permalink]
I have seen this probability question or something similar on GMAT past question.

1) I find this really strange though the multiplying of this:
outcomes * probability= probabilty

2) the problem changes if you select one ball at a time instead of two.


Can anybody explain outcomes * probability = probability

Let's say there are four different pairs of colors: blue, green, red, and yellow. You select one pair at a time.
Is the probability 1/4 for blue ?

VT
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Re: There are four pairs of shoes in a big black sack: blue, [#permalink]
This is a possible GMAT-like question. Stolyar actually writes the GMAT for ETS, he is a spy! :P

I'm too blonde and that's why I didn't see the shoes are all scrambled up and not necessary in pairs!

VT
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Re: There are four pairs of shoes in a big black sack: blue, [#permalink]
Picking first shoe 2/8
Picking the next shoe 1/7
we have such 4 pairs
4 * 2/8 * 1/7 = 1/7



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