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A box at a yard sale contains 3 different china dinner sets,

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A box at a yard sale contains 3 different china dinner sets, [#permalink] New post 17 Dec 2017, 14:09
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Question Stats:

50% (01:19) correct 50% (02:36) wrong based on 2 sessions
A box at a yard sale contains 3 different china dinner sets, each consisting of 5 plates. A customer will randomly select 2 plates to check for defects. What is the probability that the 2 plates selected will be from the same dinner set?

A. \(\frac{2}{7}\)

B. \(\frac{2}{5}\)

C. \(\frac{2}{3}\)

D. \(\frac{5}{6}\)

E. \(\frac{3}{2}\)
[Reveal] Spoiler: OA

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Re: A box at a yard sale contains 3 different china dinner sets, [#permalink] New post 18 Dec 2017, 02:21
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So here we have diffents plates. In total 15.
A customer has to choose 2 among the 15 plates.
15C2 = 15 * 14 / 2* 1 = 105 possibilities.
5C2 = 5*4/2*1 = 10
Since we have three different sets multiply this by 3.
30/105 = 0,28.... so choice A

Anyone has another quicker method?
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Re: A box at a yard sale contains 3 different china dinner sets, [#permalink] New post 18 Dec 2017, 11:51
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No, there is not.

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Re: A box at a yard sale contains 3 different china dinner sets,   [#permalink] 18 Dec 2017, 11:51
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A box at a yard sale contains 3 different china dinner sets,

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