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A computer wholesaler sells eight different computers and each is pric

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Bunuel
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A computer wholesaler sells eight different computers and each is pric [#permalink] New post   11 Dec 2014, 06:57
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Question Stats:

64% (02:03) correct 36% (02:23) wrong based on 332 sessions
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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


Source: Chili Hot GMAT­
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B
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Temurkhon
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Re: A computer wholesaler sells eight different computers and each is pric [#permalink] New post   Updated on: 12 Dec 2014, 05:53
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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] New post   11 Dec 2014, 11:23
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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] New post   12 Dec 2014, 00:43
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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] New post   19 Apr 2016, 09:21
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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] 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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GyM
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Re: A computer wholesaler sells eight different computers and each is pric [#permalink] New post   22 Jun 2018, 13:32
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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] New post   24 Jun 2018, 17:54
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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] 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?
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Re: A computer wholesaler sells eight different computers and each is pric [#permalink] New post   08 Jan 2020, 01:47
3 computers chosen from 8 computers = 8C3=8!/3!*5!=56

As the most expensive computers will be among the three chosen for display, 2C2 * 6C1=6

Required probability = Desired number of computer/ Total Computer= 6/56=3/28
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Re: A computer wholesaler sells eight different computers and each is pric [#permalink] New post   10 Jan 2020, 07:07
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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


After the two most expensive computers are chosen for the display, the number of ways to select the final computer is 6C1 = 6.

The total number of ways to select 3 computers from 8 is 8C3 = (8 x 7 x 6)/3! = 56 ways.

The overall probability is 6/56 = 3/28.

Answer: B
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Re: A computer wholesaler sells eight different computers and each is pric [#permalink] New post   16 Feb 2024, 00:20
Total outcomes= 8C3 = [(8*7*6)/(3*2*1)] =56
Favourable Outcomes= We strictly have 1 option for our first and second choice each. And 6 options to choose from for the third choice. Thus,
F/O= 1*1*6 = 6

Probability = Favourable Outcomes / Total outcomes = 6/56 = 3/28
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Re: A computer wholesaler sells eight different computers and each is pric [#permalink]
16 Feb 2024, 00:20
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