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# There are four pairs of shoes in a big black sack: blue,

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There are four pairs of shoes in a big black sack: blue, [#permalink]  15 Jul 2003, 06:30
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There are four pairs of shoes in a big black sack: blue, red, yellow, and green. Two shoes are taken at random without repetition. What is the probability of having the two of the same color?
Manager
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Shoes [#permalink]  15 Jul 2003, 07:00
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!
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2C1 x 2C1 x 2C1 x 2C1 / 8C2 = 2x2x2x2/28 = 4/7

Hypergeometric distribution applies
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Brainless

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Re: Shoes [#permalink]  15 Jul 2003, 08:04
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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I agree with mciatoo. An excellent an consise solution.
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Stolyar [#permalink]  15 Jul 2003, 15:37
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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Good Question [#permalink]  15 Jul 2003, 15:53
This is a possible GMAT-like question. Stolyar actually writes the GMAT for ETS, he is a spy!

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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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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