• Free Veritas GMAT Class
Experience Lesson 1 Live Free

Available with Beat the GMAT members only code

• Award-winning private GMAT tutoring
Register now and save up to \$200

Available with Beat the GMAT members only code

• 5 Day FREE Trial
Study Smarter, Not Harder

Available with Beat the GMAT members only code

• Free Trial & Practice Exam
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• Magoosh
Study with Magoosh GMAT prep

Available with Beat the GMAT members only code

• 1 Hour Free
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• 5-Day Free Trial
5-day free, full-access trial TTP Quant

Available with Beat the GMAT members only code

• Reach higher with Artificial Intelligence. Guaranteed
Now free for 30 days

Available with Beat the GMAT members only code

• Free Practice Test & Review
How would you score if you took the GMAT

Available with Beat the GMAT members only code

• Get 300+ Practice Questions

Available with Beat the GMAT members only code

## Tough one from Kaplan

tagged by: Brent@GMATPrepNow

This topic has 5 expert replies and 2 member replies
Joined
06 May 2013
Posted:
15 messages

#### Tough one from Kaplan

Tue Dec 10, 2013 5:36 pm
A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

### GMAT/MBA Expert

Matt@VeritasPrep GMAT Instructor
Joined
12 Sep 2012
Posted:
2637 messages
Followed by:
114 members
625
Target GMAT Score:
V51
GMAT Score:
780
Wed May 11, 2016 11:17 pm
Suppose our dogs are

AB
CD
EF
GHI

The first three pairs are obvious. Among GHI, there are three more pairs (GH, GI, HI), for six total.

Out of the 9 dogs, there are (9 choose 2) = 9*8/2 = 36 pairs.

So there are 6 littermate pairs out of 36 possible pairs, making it 1/6 that we get littermates and 5/6 that we don't.

Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!
YTarhouni Newbie | Next Rank: 10 Posts
Joined
25 Aug 2017
Posted:
8 messages
Sun Sep 03, 2017 2:20 pm
1-probability no match=1-(3*2/9*1/8+3/9*2/8)=1-(12/72)=5/6

### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
Joined
08 Dec 2008
Posted:
11405 messages
Followed by:
1229 members
5254
GMAT Score:
770
Tue Dec 10, 2013 6:00 pm
Quote:
A dog breeder has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both the selected dogs are NOT littermates?
1/6
2/9
5/6
7/9
8/9
Littermates are dogs born in the same batch (litter).

Here's the counting approach.

Let the dogs be represented by the letters A to I.
The following meets the given conditions.
- A and B are littermates
- C and D are littermates
- E and F are littermates
- G, H and I are littermates

We want to find P(selected dogs are not littermates)
Let's use the complement.
So, P(selected dogs are not littermates) = 1 - P(selected dogs are littermates)

P(selected dogs are littermates)
How many possible outcomes satisfy the condition that the two dog ARE littermates?
We have:
- A and B
- C and D
- E and F
- G and H
- G and I
- H and I
There are 6 possible outcomes.

In how many ways can we select 2 dogs from 9 dogs?
Since order doesn't matter, we can select the dogs in 9C2 ways (36 ways).

So, P(selected dogs are littermates) = 6/36 = 1/6

P(selected dogs are not littermates) = 1 - P(selected dogs are littermates)
= 1 - 1/6
= 5/6
= C

Cheers,
Brent

Aside: If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

_________________
Brent Hanneson â€“ Founder of GMATPrepNow.com
Use our video course along with

Check out the online reviews of our course
Come see all of our free resources

GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months!

### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
Joined
08 Dec 2008
Posted:
11405 messages
Followed by:
1229 members
5254
GMAT Score:
770
Tue Dec 10, 2013 6:01 pm
Quote:
A dog breeder has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both the selected dogs are NOT littermates?
1/6
2/9
5/6
7/9
8/9
Here's the probability approach.

Let the dogs be represented by the letters A to I.
The following meets the given conditions.
- A and B are littermates
- C and D are littermates
- E and F are littermates
- G, H and I are littermates

We want to find P(selected dogs are not littermates)

For the probability approach, it helps to say that one dog is selected first and the other dog is selected second.
Notice that there are two different ways in which the two dogs are NOT littermates:
#1) 1st dog is from one of the 2-dog pairings (AB, CD, or EF) and 2nd dog is not a littermate
#2) 1st dog is from the 3-dog group (GHI) and 2nd dog is not a littermate

So, . . .
P(selected dogs are not littermates) = P(1st is from a 2-dog pairing and 2nd is not a littermate OR 1st is from the 3-dog group and 2nd is not a littermate
= P(1st is from a 2-dog pairing and 2nd is not a littermate) + P(1st is from the 3-dog group and 2nd is not a littermate)
= (6/9)(7/8) + (3/9)(6/8)
= 42/72 + 18/72
= 60/72
= 5/6
= C

Cheers,
Brent

_________________
Brent Hanneson â€“ Founder of GMATPrepNow.com
Use our video course along with

Check out the online reviews of our course
Come see all of our free resources

GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months!
ash4gmat Senior | Next Rank: 100 Posts
Joined
17 Sep 2015
Posted:
57 messages
Tue May 10, 2016 11:30 pm
Brent,

You said Litter mates are born in same batch.Then how come out of 6 dogs you are making 3 pairs when the question tells out of 6 exactly 1 littermate. Same way out of 3 only 2 littermate. Please clarify.

### GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
14030 messages
Followed by:
1812 members
13060
GMAT Score:
790
Wed May 11, 2016 2:14 am
Quote:
A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

a. 1/6
b. 2/9
c. 5/6
d. 7/9
e. 8/9
Let's say that the 9 dogs are ABCDEFGHI.

6 dogs have exactly 1 littermate:
Let's say that A and B are littermates, C and D are littermates, and E and F are littermates.
This means:
A has 1 littermate (B).
B has 1 littermate (A).
C has 1 littermate (D).
D has 1 littermate (C).
E has 1 littermate (F).
F has 1 littermate (E).

3 dogs have exactly 2 littermates:
Let's say that G, H and I are all littermates of one another.
This means:
G has 2 littermates (H and I).
H has 2 littermates (G and I).
I has 2 littermates (G and H).

Total number of littermate pairs = 6:
AB, CD, EF, GH, GI, and HI.
Total number of pairs that can be formed from 9 dogs:
9C2 = 36.

P(littermate pair) = 6/36 = 1/6.
P(not a littermate pair) = 1 - 1/6 = 5/6.

If the GMAT were to use the word littermate, a definition would be offered.

_________________
Mitch Hunt
GMAT Private Tutor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.
Available for tutoring in NYC and long-distance.

Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.

### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
Joined
23 Jun 2013
Posted:
9305 messages
Followed by:
478 members
2867
GMAT Score:
800
Wed May 11, 2016 8:49 am

This is a quirky probability question that requires that you keep track of a number of details. There are a few ways to do the math; here's how I would approach it:

We're told that there are 9 dogs, 6 of them have 1 litter mate and 3 of them have 2 litter mates. ALL of these dogs are contained within the group of 9 dogs.

So, let's call the dogs:
1 litter mate:
A & B
C & D
E & F

2 litter mates:
G, H and I

The question asks for the probability that 2 dogs, selected at random, are NOT litter mates.
I'm going to do the math in 2 calculations:

If the first dog is one of the "1 litter mate" dogs:
(6/9)
then on the next dog, (7/8) are NOT litter mates:
(6/9)(7/8) = 42/72

If the first dog is one of the "2 litter mate" dogs:
(3/9)
then on the next dog, (6/8) are NOT litter mates:
(3/9)(6/8) = 18/72

In TOTAL, (42/72) + (18/72) = 60/72 = 5/6

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

_________________
Contact Rich at Rich.C@empowergmat.com

### Best Conversation Starters

1 lheiannie07 84 topics
2 LUANDATO 60 topics
3 ardz24 56 topics
4 M7MBA 50 topics
5 AAPL 44 topics
See More Top Beat The GMAT Members...

### Most Active Experts

1 Rich.C@EMPOWERgma...

EMPOWERgmat

136 posts
2 GMATGuruNY

The Princeton Review Teacher

133 posts
3 Brent@GMATPrepNow

GMAT Prep Now Teacher

130 posts
4 Scott@TargetTestPrep

Target Test Prep

118 posts
5 Jeff@TargetTestPrep

Target Test Prep

114 posts
See More Top Beat The GMAT Experts