Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 May 2017, 18:12

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

When choose 3 out of 8 books, the probability of the most ex

Author Message
TAGS:

Hide Tags

Manager
Joined: 02 Apr 2006
Posts: 93
Followers: 1

Kudos [?]: 42 [0], given: 0

When choose 3 out of 8 books, the probability of the most ex [#permalink]

Show Tags

10 Nov 2007, 20:39
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (01:38) wrong based on 2 sessions

HideShow timer Statistics

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Manager
Joined: 02 Apr 2006
Posts: 93
Followers: 1

Kudos [?]: 42 [0], given: 0

Re: (probability) 3 out of 8, the most expensive two include [#permalink]

Show Tags

10 Nov 2007, 21:04
Vemuri wrote:
pretttyune wrote:

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Is the question complete?

yes... the question is .. In choosing 3 out of 8 books, the probability of the most expensive two books must be included?
VP
Joined: 22 Nov 2007
Posts: 1083
Followers: 8

Kudos [?]: 567 [0], given: 0

Show Tags

08 Jan 2008, 22:58
walker wrote:
P=6C1/8C3=6*3*2/(8*7*6)=3/28

it is: consider the most expensive 2 books....than we have 6 slots free to combine the books. the total number of combs is 8C3 because we have a total of 8 books to combine in groups of three...prob=6/8*7=3/28
SVP
Joined: 28 Dec 2005
Posts: 1561
Followers: 3

Kudos [?]: 158 [0], given: 2

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

Show Tags

09 Jan 2008, 06:44
i thought about it like this.

probability = desired outcomes / total out comes

where desired outcomes is the number of outcomes of picking 3 out of 8 books, but the two most expensive have to be in there, so ive got (1C1)*(1C1)*(6C1) / 8C3 = 6/56 = 3/28
CEO
Joined: 21 Jan 2007
Posts: 2745
Location: New York City
Followers: 11

Kudos [?]: 942 [0], given: 4

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

Show Tags

09 Jan 2008, 10:01
i did it slightly differently.

(2/8 *1/7 * 6/6) * 3c1 = 3/28
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Intern
Joined: 07 Nov 2006
Posts: 14
Followers: 0

Kudos [?]: 6 [0], given: 0

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

Show Tags

25 Jan 2008, 05:19
walker wrote:
walker

Walker pls elucidate why not 2!6c1/8c3
CEO
Joined: 17 Nov 2007
Posts: 3586
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3982 [0], given: 360

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

Show Tags

25 Jan 2008, 05:35
AlexBon wrote:
walker wrote:
walker

Walker pls elucidate why not 2!6c1/8c3

$$p=\frac{C^2_2*C^6_1}{C^8_3}=\frac{1*6*3*2}{8*7*6}=\frac{3}{28}$$

or

$$p=\frac{C^2_2*C^6_1*P^3_3}{P^8_3}=\frac{1*6*3*2}{8*7*6}=\frac{3}{28}$$

or

$$p=\frac28*(\frac17*\frac66+\frac67*\frac16)+\frac68*\frac27*\frac16=\frac{3*12}{8*7*6}=\frac{3}{28}$$

I think 2! is permutation: $$P^2_2$$. But you use $$C^8_3$$ for all variants that mean ABC and BAC are the same variant. If we distinguish between ABC and BAC, we use $$P^8_3$$ and $$P^3_3$$

Hope this help.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 150 [0], given: 3

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

Show Tags

28 Sep 2009, 10:50
Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Soln: 6C1/8C3
Senior Manager
Joined: 22 Dec 2009
Posts: 359
Followers: 11

Kudos [?]: 389 [0], given: 47

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

Show Tags

17 Feb 2010, 03:44
pretttyune wrote:

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Out of 8 .. 2 books (which are most exp) should be considered in your selection.

No of ways the 2 Exp books can be selected = 2c2 = 1

remaining we need to choose one more book from the 6 left over books = 6c1 = 6

Total ways of selecting 3 books out of 8 = 8c3

Prob = 1X6 / 8c3 = 3/28
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Intern
Joined: 02 May 2013
Posts: 25
WE: Engineering (Aerospace and Defense)
Followers: 1

Kudos [?]: 49 [0], given: 16

Re: (probability) 3 out of 8, the most expensive two included... [#permalink]

Show Tags

05 Aug 2013, 21:14
Probability of selecting most expensive out of 8 =1/8
Probability of selecting next most expensive out of 7 =1/7
And out of 6 left out can select any one out of 6 = 6c1/6c1
now probability of selecting 2 most expensive = 1/8*1/7*1*3!
=3/8(3! is because of arrangements)
Re: (probability) 3 out of 8, the most expensive two included...   [#permalink] 05 Aug 2013, 21:14
Similar topics Replies Last post
Similar
Topics:
3 A book store has received 8 different books, of which 3/8 are novels, 3 24 May 2017, 23:39
9 What is the remainder when 8^1+ 8^2+ 8^3……. + 8^15 is divided by 6 9 04 Dec 2016, 00:34
2 What is the probability that you will get a sum of 8 when 3 12 Dec 2015, 03:49
4 If n is a positive integer, what is the remainder when 3^(8n 8 18 Feb 2014, 02:06
9 If n is a positive integer, what is the remainder when 3^(8n 8 07 Jun 2014, 04:24
Display posts from previous: Sort by