Probability

Expert replies

abhirup1711: Senior | Next Rank: 100 Posts; Posts: 66; Joined: Wed Jul 13, 2011 2:27 pmFollowed by: 2 members

by abhirup1711 » Wed Jun 12, 2013 6:26 am

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

Like this post

Source: Beat The GMAT — Problem Solving | Wed Jun 12, 2013 6:26 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Jun 12, 2013 7:06 am

abhirup1711 wrote: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
= [spoiler]5/6 [/spoiler]
= C

Cheers,
Brent

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

Brent Hanneson - Creator of GMATPrepNow.com

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Jun 12, 2013 7:16 am

I posted a solution to this problem here:
https://www.beatthegmat.com/dog-breeder-t134956.html

Check here for a similar problem:
https://www.beatthegmat.com/siblings-t66114.html

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

Like this post

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Jun 12, 2013 7:30 am

abhirup1711 wrote: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
=[spoiler] 5/6[/spoiler]
= C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review