(probability) 3 out of 8, the most expensive two included...

Oldest

Best Reply

Author

Message

TAGS:

Manager

Joined: 02 Apr 2006

Posts: 94

Followers: 1

Kudos [?]: 1 [0], given: 0

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

10 Nov 2007, 20:39

00:00

Question Stats:

0% (00:00) correct

100% (01:38) wrong

based on 0 sessions

Please help me to understand this... I am always terrified when i face PROBABILITY questions...

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

Please helpe me kindly and thank you in advance...

Manager

Joined: 02 Apr 2006

Posts: 94

Followers: 1

Kudos [?]: 1 [0], given: 0

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

10 Nov 2007, 21:04

Vemuri wrote:

pretttyune wrote:

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?

Joined: 22 Nov 2007

Posts: 1108

Followers: 6

Kudos [?]: 80 [0], given: 0

Re: [#permalink]

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: 1612

Followers: 1

Kudos [?]: 53 [0], given: 2

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

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: 2797

Location: New York City

Followers: 5

Kudos [?]: 132 [0], given: 4

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

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: 15

Followers: 0

Kudos [?]: 1 [0], given: 0

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

25 Jan 2008, 05:19

walker wrote:

walker

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

CEO

Joined: 17 Nov 2007

Posts: 3591

Concentration: Entrepreneurship, Other

Schools: Chicago (Booth) - Class of 2011

GMAT 1: 750 Q50 V40

Followers: 230

Kudos [?]: 1298 [0], given: 346

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

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.
_________________

iPhone/iPod/iPad: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
Android: GMAT ToolKit (NEW!). POLL: What tool do you need next?
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.

Find out what's new at GMAT Club - latest features and updates

Manager

Joined: 27 Oct 2008

Posts: 188

Followers: 1

Kudos [?]: 42 [0], given: 3

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

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: 368

Followers: 9

Kudos [?]: 136 [0], given: 47

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

17 Feb 2010, 03:44

pretttyune wrote:

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!! :beer

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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

Re: (probability) 3 out of 8, the most expensive two included... [#permalink] 17 Feb 2010, 03:44

		Similar topics	Author	Replies	Last post
Similar Topics:		Out of 8 employees including Bob and Mary, a team of 4	Caas	5	12 May 2007, 04:18
		Two pieces of fruit are selected out of a group of 8 pieces	zisis	12	29 Sep 2010, 15:39
		What is the probability that two siblings, ages 6 and 8	iNumbv	1	02 May 2012, 14:27
	3	Two pieces of fruit are selected out of a group of 8 pieces	Pansi	7	09 Nov 2012, 01:34
	4	The probability that a target will be shot two out of two	SravnaTestPrep	8	17 Mar 2013, 18:53

(probability) 3 out of 8, the most expensive two included...

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.

GMAT Courses

Admissions Consulting

(probability) 3 out of 8, the most expensive two included...

(probability) 3 out of 8, the most expensive two included...

Main navigation

GMAT Resources

Partners

MBA Resources