A group of 3 integers is to be selected one after the other, : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 06:12

### 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

# A group of 3 integers is to be selected one after the other,

Author Message
TAGS:

### Hide Tags

Intern
Joined: 24 Nov 2011
Posts: 2
Followers: 0

Kudos [?]: 3 [1] , given: 0

A group of 3 integers is to be selected one after the other, [#permalink]

### Show Tags

24 Nov 2011, 10:45
1
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

68% (02:56) correct 32% (02:55) wrong based on 114 sessions

### HideShow timer Statistics

List L: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

A group of 3 integers is to be selected one after the other, and at random and without replacement, from list L above. What is the probability that the 3 integers selected are not 3 consecutive integers?

A. 3/5
B. 7/10
C. 3/4
D. 4/5
E. 14/15

One way to do this is to count all the consecutive combinations and divide that by 10C3, subtract the whole thing from 1. But is there a more systematic and efficient way to do this problem? kaplan books only provide brute force solutions without any formulas. thanks.
[Reveal] Spoiler: OA
 Kaplan GMAT Prep Discount Codes EMPOWERgmat Discount Codes Economist GMAT Tutor Discount Codes
Manager
Joined: 29 Oct 2011
Posts: 184
Concentration: General Management, Technology
Schools: Sloan '16 (D)
GMAT 1: 760 Q49 V44
GPA: 3.76
Followers: 10

Kudos [?]: 138 [1] , given: 19

Re: The most efficient way to do this probability question [#permalink]

### Show Tags

24 Nov 2011, 11:36
1
KUDOS
The method you suggested is the simplest. It's very easy to count the number of series 3 consecutive integers, especially since you don't need to do permutations. There are a total of 8 such series (1,2,3), (2,3,4)...(8,9,10).

1 - 8/10C3 = 1 - 1/15 = 14/15 E
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2138

Kudos [?]: 13691 [2] , given: 222

Re: The most efficient way to do this probability question [#permalink]

### Show Tags

24 Nov 2011, 21:54
2
KUDOS
Expert's post
topspin330 wrote:
List L: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

A group of 3 integers is to be selected one after the other, and at random and without replacement, from list L above. What is the probability that the 3 integers selected are not 3 consecutive integers?

a - 3/5
b - 7/10
c - 3/4
d - 4/5
e - 14/15

One way to do this is to count all the consecutive combinations and divide that by 10C3, subtract the whole thing from 1. But is there a more systematic and efficient way to do this problem? kaplan books only provide brute force solutions without any formulas. thanks.

There is a little bit of ambiguity in this question. When you read it, you wonder whether the order of selection is important i.e. should the numbers be selected in a consecutive manner i.e. are they looking for the probability of selecting 2, then 3 and then 4 or is this selection made in any other order e.g. 3, then 2 and then 4 fine too.
You need to guess from the answer options that the order of selection is not important. In that case, the solution provided above is the best and fastest method.

In case the order of selection is important, the required probability = (8/10)*(1/9)*(1/8) = 1/90
i.e. you can select any of the first 8 integers first. Now both the second and the third pick are defined e.g. if you select 4 on your first pick, you need to select 5 next (probability of that = 1/9 since there are total 9 numbers left) and then you need to select 6 (probability of that = 1/8 since there are total 8 numbers left)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: Essaying Joined: 27 May 2010 Posts: 153 Location: Ghana Concentration: Finance, Finance Schools: Cambridge GMAT 1: 690 Q47 V37 GPA: 3.9 WE: Accounting (Education) Followers: 1 Kudos [?]: 9 [0], given: 8 Re: The most efficient way to do this probability question [#permalink] ### Show Tags 25 Nov 2011, 03:26 Went by counting the possible 3 consecutive set = 8 8/10C3 = 8/120 1/15 15/15-1/15 = 14/15 Manager Joined: 16 Dec 2009 Posts: 75 GMAT 1: 680 Q49 V33 WE: Information Technology (Commercial Banking) Followers: 1 Kudos [?]: 39 [0], given: 11 Re: The most efficient way to do this probability question [#permalink] ### Show Tags 25 Nov 2011, 04:39 As Karishma said , there is an ambiguity in the question. I guess one way to find the answer is to approach in both ways and check if only one answer is present among the answer choices. If order of the consecutive numbers is not important , then answer would be 1- {8/10C3} or 1 - {8/120} or 14/15 If order of the consecutive numbers is important , then answer would be 1- {(8*3!)/10C3} or 1- {48/120} or 3/5. In this case , both the answers are present in the answer choices ( ), so question needs to be little more clear. I don't think this kind of unclear question would ever come in actual GMAT . _________________ If Electricity comes from Electrons , Does Morality come from Morons ?? If you find my post useful ... then please give me kudos ...... h(n) defined as product of even integers from 2 to n Number N divided by D leaves remainder R Ultimate list of MBA scholarships for international applicants Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2138 Kudos [?]: 13691 [0], given: 222 Re: The most efficient way to do this probability question [#permalink] ### Show Tags 27 Nov 2011, 22:44 Responding to a pm: Wy do you restrict the first choice to 8 options?, the way I understand this problem you could pick any of the numbres 1 to 10 and then the next two would have constraints. For example if my first pick is 10, my second 8 and my last pick is 9, wouldn't that set comply with being a set of consecutive numbers? The point is 'does the order of picking numbers matter'. If only the end result is important i.e. you get three numbers which are consecutive, then the answer is different but if the order of selecting numbers is important too i.e. after 8, I must pick 9 and then 10, then the answer is different. The question would have been clearer if they had mentioned that three numbers are picked simultaneously. Then the order doesn't matter. When they say that the numbers are picked one after the other, it gets you thinking if the order of selection is important too. Anyway, the intention of the original question is what you suggested. I offered a different spin on it. In fact, there are two different spins on the question. When I restrict the first number to one of 1-8, I am assuming that they want the probability of 3 consecutive numbers picked in increasing order. If you want to find the probability of 3 consecutive number picked in either increasing or decreasing order (say you pick 8, 9, 10 OR you pick 8, 7, 6 in that order), you do not need to have this restriction but you do need different cases. If you pick 1/2/9/10, the next two numbers can be chosen in only 1 way each. If you pick any other number, you can choose the next number in 2 ways and the third number in one way. Convert this logic into math and find out what you get. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Manager
Joined: 25 May 2011
Posts: 156
Followers: 2

Kudos [?]: 60 [0], given: 71

Re: The most efficient way to do this probability question [#permalink]

### Show Tags

13 Dec 2011, 02:44
The question has to be changed. It's not clear.
Manager
Joined: 14 Dec 2011
Posts: 64
GMAT 1: 630 Q48 V29
GMAT 2: 690 Q48 V37
Followers: 1

Kudos [?]: 34 [0], given: 24

Re: The most efficient way to do this probability question [#permalink]

### Show Tags

27 Dec 2011, 04:40
I just thought of another way to figure out the problem:

1.) Select any number: P1 10/10
2.) Select any number next to the first one: P2 2/9
3.) Select and number next to the first or second chosen one: P3 2/8

P1 * P2 * P3 = 1/18

1 - 1/18 = 17/18 .... this is closest to E. The only problem is that I can also chose numbers 1 and 10 which don't have two neighboring numbers.

Taking this into account will make it too difficult and long-lasting (at least for me)
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13499
Followers: 577

Kudos [?]: 163 [0], given: 0

Re: The most efficient way to do this probability question [#permalink]

### Show Tags

07 Dec 2013, 00:14
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13499
Followers: 577

Kudos [?]: 163 [0], given: 0

Re: A group of 3 integers is to be selected one after the other, [#permalink]

### Show Tags

12 Oct 2015, 02:40
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.
_________________
SVP
Joined: 17 Jul 2014
Posts: 2209
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '19 (S)
GMAT 1: 560 Q42 V26
GMAT 2: 550 Q39 V27
GMAT 3: 560 Q43 V24
GMAT 4: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Followers: 20

Kudos [?]: 271 [0], given: 140

Re: A group of 3 integers is to be selected one after the other, [#permalink]

### Show Tags

09 Dec 2015, 19:08
I solved it using permutations:
there are 10C3 number of ways to select 3 numbers, or 120 ways.
I struggled to find the number of answers using probability and counting...but later, I just simply put 3 numbers in a "big" number, 3 numbers of course mean 3 consecutive numbers. Now, we have one "big" numbers and 7 other numbers. in how many ways can we select the big number? 8C1 or 8 ways. Thus, the probability of not being consecutive is (120-8)/120, which is 14/15
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 8314
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Followers: 381

Kudos [?]: 2468 [0], given: 163

Re: A group of 3 integers is to be selected one after the other, [#permalink]

### Show Tags

10 Dec 2015, 17:42
Hi mvictor,

Determining that there were 120 ways to select 3 numbers from that group was the perfect first step. Next, you just have to figure out which combinations of 3 involve three consecutive integers. Thankfully, it's not that hard - and you can list them out:

123
234
345
456
567
678
789
89&10

Thus there are 8 ways (out of 120) that ARE three consecutive integers and the other 112/120 are NOT three consecutive integers.

112/120 = 14/15

[Reveal] Spoiler:
E

GMAT assassins aren't born, they're made,
Rich
_________________

# Rich Cohen

Co-Founder & GMAT Assassin

# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** SVP Joined: 17 Jul 2014 Posts: 2209 Location: United States (IL) Concentration: Finance, Economics Schools: Stanford '19 (S) GMAT 1: 560 Q42 V26 GMAT 2: 550 Q39 V27 GMAT 3: 560 Q43 V24 GMAT 4: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) Followers: 20 Kudos [?]: 271 [0], given: 140 A group of 3 integers is to be selected one after the other, [#permalink] ### Show Tags 10 Dec 2015, 19:03 EMPOWERgmatRichC wrote: Hi mvictor, Determining that there were 120 ways to select 3 numbers from that group was the perfect first step. Next, you just have to figure out which combinations of 3 involve three consecutive integers. Thankfully, it's not that hard - and you can list them out: 123 234 345 456 567 678 789 89&10 Thus there are 8 ways (out of 120) that ARE three consecutive integers and the other 112/120 are NOT three consecutive integers. 112/120 = 14/15 Final Answer: [Reveal] Spoiler: E GMAT assassins aren't born, they're made, Rich Thanks Rich for explanation. I usually have problems with combinatorics problems, especially when some restrictions are present. For example, in how many ways can we select 3 numbers so that the sum of these 3 numbers is not divisible by 2... Knowing that it is one of my greatest weaknesses, I sometimes overthink, and do not follow the "easy" step - actually counting the possibilities.. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 8314 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 381 Kudos [?]: 2468 [0], given: 163 Re: A group of 3 integers is to be selected one after the other, [#permalink] ### Show Tags 11 Dec 2015, 11:05 Hi mvictor, Many of these skills develop with time and practice. When these types of probability/permutation/combination questions show up on Test Day, the solution often 'leans' one way or the other, so it helps to think about which of two things would be easier to calculate: 1) What we want (what the question asks for) 2) What we don't want (the options that DON'T fit what the question asks for, so that we can subtract that number from the total to get the answer to the question). Regardless, these types of questions are NOT worth a lot of points on Test Day. You shouldn't be spending too much time on these types of prompts if you're missing out on lots of points in the BIG Quant categories. GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

60-point improvement guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Re: A group of 3 integers is to be selected one after the other,   [#permalink] 11 Dec 2015, 11:05
Similar topics Replies Last post
Similar
Topics:
1 If a 3-digit integer is selected at random from the integers 100 throu 3 26 Nov 2015, 04:47
1 There are 5 competitors in a race (John, mary and 3 other ones) in whi 1 15 Jul 2015, 17:35
3 In how many ways can a group of 3 people be selected 5 09 Nov 2012, 09:59
3 What is the probability that one of the two integers randomly selected 6 09 May 2010, 12:09
If a 3-digit integer is selected at random from the 8 11 Nov 2007, 14:29
Display posts from previous: Sort by