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A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl

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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]

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Joy111 wrote:
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


Here's an approach using probability rules:

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
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]

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New post 17 Oct 2016, 08:10
GMATPrepNow wrote:
Joy111 wrote:
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


Here's an approach using probability rules:

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


Exactly my point in the previous post, but you explained it with more detail.
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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl [#permalink]

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New post 07 Dec 2017, 14:36
Hi All,

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:
[Reveal] Spoiler:
C


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Re: A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl   [#permalink] 07 Dec 2017, 14:36

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