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A computer wholesaler sells eight different computers and each is pric

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A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 11 Dec 2014, 06:57
4
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A
B
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D
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  55% (hard)

Question Stats:

60% (01:39) correct 40% (01:51) wrong based on 207 sessions

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Tough and Tricky questions: Combinations.



A computer wholesaler sells eight different computers and each is priced differently. If the wholesaler chooses three computers for display at a trade show, what is the probability (all things being equal) that the two most expensive computers will be among the three chosen for display?

A) 15/56
B) 3/28
C) 1/28
D) 1/56
E) 1/168

Kudos for a correct solution.

Source: Chili Hot GMAT

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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post Updated on: 12 Dec 2014, 05:53
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3
Direct probability:

2/8*1/7*1=1/28

we have 3!/2!*1!=3 such cases

(1/28)*3=3/28

Reverse probability:

Not desirable cases when we have all three not highest price or only one in highest price

6/8*5/7*4/6=5/14

2/8*6/7*5/6=(5/28)*3 cases=15/28

1 - (5/14 + 15/28) = 1 - 25/28=3/28

Combination approach:

denominator (all possible cases): 8C3=8!/3!*5!=56

numerator (desirable cases): 2C2*6C1=6

6/56=3/28

Reverse combination:

denimonator (all cases): 8C3=56

numerator (undesirable cases): 6C3 + (2C1*6C2) = 50

1 - 50/56=6/56=3/28


B


in hope to get Kudos from Bunuel) and you, guys

Originally posted by Temurkhon on 11 Dec 2014, 22:42.
Last edited by Temurkhon on 12 Dec 2014, 05:53, edited 3 times in total.
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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 11 Dec 2014, 11:23
2
Bunuel wrote:

Tough and Tricky questions: Combinations.



A computer wholesaler sells eight different computers and each is priced differently. If the wholesaler chooses three computers for display at a trade show, what is the probability (all things being equal) that the two most expensive computers will be among the three chosen for display?

A) 15/56
B) 3/28
C) 1/28
D) 1/56
E) 1/168

Kudos for a correct solution.

Source: Chili Hot GMAT


8!/3!*5!=56
1( 1 way of picking top 2 comps in pricing list) ∗ 6 ( ypu can pick third comp in 6 ways as there are 6 more left)=6
and then favourite outcomes / total number of outcomes => 6/56 = 3/28, so B?
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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 12 Dec 2014, 00:43
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1
Since, two of the choices are prefixed, we are free to choose 1 from the rest of the 6 avilable. so 6C1 is the numerator.

Total no of ways we can choose 3 from 8 is 8c3, which is the denominator.
So, the probability: 6c1/8c3= 3/28 ans is B.
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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 19 Apr 2016, 09:21
1
Bunuel wrote:

Tough and Tricky questions: Combinations.



A computer wholesaler sells eight different computers and each is priced differently. If the wholesaler chooses three computers for display at a trade show, what is the probability (all things being equal) that the two most expensive computers will be among the three chosen for display?

A) 15/56
B) 3/28
C) 1/28
D) 1/56
E) 1/168

Kudos for a correct solution.

Source: Chili Hot GMAT


Probability = P(most expensive) * P(2nd most expensive)*P(any other) *3! = 1/8*1/7*6/6*3! =3/28
So answer B

Please suggest if the approach is correct.!!!
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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 22 Jun 2018, 12:53
Let the two most expensive computers be A & B.

Since three are chosen out of 8 & two of the three have to be A & B, we get selection as A B _

Required Probability = 1/8*1/7*6/6*3! = 6/56 = 3/28

Answer B.



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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 22 Jun 2018, 13:32
Total possibilities = 8C3 = 56

Out of the 3 to be chosen. Let's say each computer has a name. A,B,C,D,E,F,G,H

And let's say that G and H are the most expensive.

Now - think of the slot method for the 3 computers to be chosen. _ _ _

To ensure we select both of the most expensive computers (G and H), we populate the first slot with H and second with G.

H G _

The third slot can be filled with any of the remaining 6 computers. Now you have the following:

1 * 1 * 6 = 6 (H placed int the first slot, G placed in the second slot, and any of the 6 remaining computers placed in the 3rd slot.

6/56 = 3/28

Answer B
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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 24 Jun 2018, 17:54
Bunuel wrote:
[header3]

A computer wholesaler sells eight different computers and each is priced differently. If the wholesaler chooses three computers for display at a trade show, what is the probability (all things being equal) that the two most expensive computers will be among the three chosen for display?

A) 15/56
B) 3/28
C) 1/28
D) 1/56
E) 1/168


Three computers can be selected from 8 computers in 8C3 = (8 x 7 x 6)/3! = 8 x 7 = 56 ways.

The two most expensive computers can be selected in 2C2 = 1 way.

The third computer can be selected from the remaining 6 computers in 6C1 = 6 ways.

Thus, the probability is (2C2 x 6C1)/8C3 = (1 x 6)/56 = 3/28.

Answer: B
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Re: A computer wholesaler sells eight different computers and each is pric  [#permalink]

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New post 26 Jun 2018, 17:26
Temurkhon wrote:
Direct probability:

2/8*1/7*1=1/28

we have 3!/2!*1!=3 such cases

(1/28)*3=3/28

Reverse probability:

Not desirable cases when we have all three not highest price or only one in highest price

6/8*5/7*4/6=5/14

2/8*6/7*5/6=(5/28)*3 cases=15/28

1 - (5/14 + 15/28) = 1 - 25/28=3/28

Combination approach:

denominator (all possible cases): 8C3=8!/3!*5!=56

numerator (desirable cases): 2C2*6C1=6

6/56=3/28

Reverse combination:

denimonator (all cases): 8C3=56

numerator (undesirable cases): 6C3 + (2C1*6C2) = 50

1 - 50/56=6/56=3/28


B


in hope to get Kudos from Bunuel) and you, guys




I was wondering how did you determine that the number of cases were 3, why 3?
Re: A computer wholesaler sells eight different computers and each is pric &nbs [#permalink] 26 Jun 2018, 17:26
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