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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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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] New post 05 Jan 2019, 21:14
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Popo wrote:
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?


I have one quicker way

the probability of selecting any plate is 1, the probability of selecting a matching plate after that is 4/14
so 1*4/14= 4/14= 2/7
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Re: A box at a yard sale contains 3 different china dinner sets, [#permalink] New post 05 Jan 2019, 21:32
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I have two methods for this.

the longer way is:
total number of combinations is 15c2
the total number of ways to select 5 plates is 5c2
the total number of ways to select 3 different sets is 3c2

then
(5c2*3c2)/15c2 = 10*3/105= 30/105= 2/7

the shorter way is..

the probability of selecting any plate is 1
the probability of selecting a matching plate out of the 14 other plates is 4/14
1*4/14=2/7
the answer is A
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Re: A box at a yard sale contains 3 different china dinner sets, [#permalink] New post 06 Jan 2019, 08:49
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Carcass wrote:
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. \(\frac{3}{2}\)



Another way..
Any of the 15 can be picked up in the first go. so P = \(\frac{15}{15}=1\)
Next can be any of the remaining 10 to ensure they are from different sets so P = \(\frac{10}{14}\)
Over all prob = \(1*\frac{10}{14}=\frac{5}{7}\)

Thus required probability = 1- \frac{5}{7}=\frac{2}{7}[/m]
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Re: A box at a yard sale contains 3 different china dinner sets, [#permalink] New post 06 Jan 2019, 10:34
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Carcass wrote:
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. \(\frac{3}{2}\)


Solution:

Selecting 2 plates from all the plates ( 3 different sets and each having 5 plates = 3x5 = 15) => 15c2 = (15x14)/2 = 15 x 7

Selecting 2 plates from same set = Selecting 1 set from 3 sets (3c1) and Selected 2 plates from the set selected (5c2).

i.e; 3c1 x 5c2 = 30

Probability = (3c1 x 5c2) / 15c2
= 30/ (15 x 7)
= 2/7
Re: A box at a yard sale contains 3 different china dinner sets,   [#permalink] 06 Jan 2019, 10:34
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