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
OA is c
A dog breeder
This topic has expert replies
- conquistador
- Master | Next Rank: 500 Posts
- Posts: 266
- Joined: Fri Sep 19, 2014 4:00 am
- Thanked: 4 times
- Followed by:1 members
GMAT/MBA Expert
- [email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi Mechmeera,
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
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Let's say that the 9 dogs are ABCDEFGHI.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
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 each other.
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.
The correct answer is C.
If the GMAT were to use the word littermate, a definition would be offered.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Here's an approach using probability rules: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?
1/6
2/9
5/6
7/9
8/9
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
=[spoiler] 5/6[/spoiler]
= C
Cheers,
Brent