GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 03 Aug 2020, 10:30

### 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

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

Author Message
Manager
Joined: 02 Apr 2006
Posts: 75
When choose 3 out of 8 books, the probability of the most ex  [#permalink]

### Show Tags

10 Nov 2007, 19:39
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: 75
Re: (probability) 3 out of 8, the most expensive two include  [#permalink]

### Show Tags

10 Nov 2007, 20:04
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?
Director
Joined: 22 Nov 2007
Posts: 802

### Show Tags

08 Jan 2008, 21:58
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
VP
Joined: 28 Dec 2005
Posts: 1027
Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

### Show Tags

09 Jan 2008, 05:44
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
SVP
Joined: 21 Jan 2007
Posts: 1856
Location: New York City
Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

### Show Tags

09 Jan 2008, 09:01
1
i did it slightly differently.

(2/8 *1/7 * 6/6) * 3c1 = 3/28
Intern
Joined: 07 Nov 2006
Posts: 8
Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

### Show Tags

25 Jan 2008, 04:19
walker wrote:
walker

Walker pls elucidate why not 2!6c1/8c3
CEO
Joined: 17 Nov 2007
Posts: 2913
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

### Show Tags

25 Jan 2008, 04:35
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 Club Forum 2020 | GMAT ToolKit 2 (iOS) - The OFFICIAL GMAT CLUB PREP APPs, must-have apps especially if you aim at 700+
Manager
Joined: 27 Oct 2008
Posts: 125
Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

### Show Tags

28 Sep 2009, 09:50
Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Soln: 6C1/8C3
Manager
Joined: 22 Dec 2009
Posts: 225
Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

### Show Tags

17 Feb 2010, 02:44
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
Intern
Joined: 02 May 2013
Posts: 18
WE: Engineering (Aerospace and Defense)
Re: (probability) 3 out of 8, the most expensive two included...  [#permalink]

### Show Tags

05 Aug 2013, 20:14
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)
Non-Human User
Joined: 09 Sep 2013
Posts: 15595
Re: When choose 3 out of 8 books, the probability of the most ex  [#permalink]

### Show Tags

15 Jan 2019, 01:44
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: When choose 3 out of 8 books, the probability of the most ex   [#permalink] 15 Jan 2019, 01:44