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

Question

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

A. 1/6
B. 2/9
C. 5/6
D. 7/9
E. 8/9

Bunuel · Accepted Answer

Since there are 6 dogs with exactly 1 littermate each, then we have three pairs of littermates: (1-2), (3-4), (5-6); Since there are 3 dogs with exactly 2 littermates each, then we have one triple of littermates: (7-8-9). Let's find the probability of the opposite event, so the probability that both selected dogs ARE littermates, and subtract it from 1. We can select either (1-2), (3-4), or (5-6) so 3 ways; We can select any two from (7-8-9) so C^2_3=3 ways; So, total ways to select 2 littermates out of these 9 dogs is 3+3=6; Total # of ways to select 2 dogs out of 9 is C^2_9=36 ; P=1-6/36=5/6. Answer: C. Similar question to practice: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html Hope it helps.

KarishmaB · Answer

There is a very similar question here: probability-siblings-in-the-room-80722.html#p922063 You can use the exact same logic to arrive at the answer. Try to take one line at a time to reason it out. Everything will easily fall in place. I will tell you what I mean. I read the first line: "In a room filled with 9 breeding dogs, 6 have exactly 1 littermate and" I stop here. 6 dogs have exactly 1 littermate. So I say to myself, "Ok. A is there and it has a littermate B. Wait, this means that automatically B has the littermate A. They cannot have any other littermate since both should have only one littermate each. Then out of 6 dogs , 2 are already accounted for. Then in the same way, there must be C and its littermate D and there must also be E and its littermate F". The important thing to realize is that 'littermate' relation is symmetric. If A is littermate of B, B is also the littermate of A. Next line of the question: "3 of the dogs have exactly 2 littermates" Now I think, "There is G and it has two littermates H and J. So automatically H and J both have 2 littermates each too." This gives me following littermates: A - B C - D E - F G - H - I Now I need to pick 2 dogs such that they are not littermates. I can do it in 2 ways. I can either find the number of ways of picking littermates or number of ways of picking 'non littermates'. Number of ways of picking littermates is certainly easier - I can pick A-B or C-D or E-F or 2 of G-H-I in 3 ways (G-H or H-I or G-I) so there are a total of 6 ways of picking littermates. Total ways of picking 2 dogs is 9C2 = 36 ways Out of these 36 ways, 6 ways are there to pick 2 littermates so other 30 ways must be of picking 'non littermates'. Required probability = 30/36 = 5/6

KarishmaB · Answer

There are a total of 9 dogs - 3 have exactly 2 litter mates. 6 have only 1 litter mate.
Say A has two litter mates B and C. Then B also has two litter mates - A and C and C also has two litter mates A and B. There will be only one such group of 3 dogs where each has exactly 2 litter mates.
Of the remaining 6, say D's litter mate is E. Then E's litter mate is D. Neither of them will have any other litter mate.
Similarly, F's litter mate is G and G's litter mate is F.
H's litter mate is I and I's litter mate is H.

A, B, C
D, E
F, G
H, I

In how many ways can we select two dogs such that they ARE litter mates?
You can do it in two ways:

Finding number of combinations:
We can select D, E or F, G or H, I i.e. 3 ways.
Of A, B, C, we can select any two in 3C2 = 3 ways.
Total ways of selecting 2 litter mates is 3+3 = 6

Total ways of selecting 2 dogs out of 9 is 9C2 = 36

Probability of selecting 2 dogs such that they are litter mates = 6/36 = 1/6
Probability of selecting 2 dogs such that they are not litter mates = 1 - 1/6 = 5/6

Directly using probability:

Probability of selecting a dog out of D, E, F, G, H and I = 6/9 = 2/3
Probability of selecting his litter mate = 1/8

Probability of selecting a dog out of A, B and C = 3/9 = 1/3
Probability of selecting his litter mate = 2/8 = 1/4

Probability of selecting 2 litter mates = 2/3 *1/8 + 1/3*1/4 = 1/6
Probability of selecting non litter mates = 1 - 1/6 = 5/6

garimavyas · Answer

what is the meaning of a litter mate btw and what if on the test day i get this kind of term which i dont understand like a dogs littermate .

KarishmaB · Answer

Yeah, I wasn't thrilled when I came across this word either. But it isn't hard to guess who a litter mate is. A litter mate would be a dog born in the same litter. I really doubt you will get any ambiguous terminology on the real test.

LalaB · Answer

1) select no littermates from Group 1 (1-2), (3-4), (5-6) 
6*2=12 (i.e. 1-3; 1-4; 1-5;1-6;2-3;2-4;2-5;2-6;3-5;3-6;4-5;4-6)

2) select no littermates from Group 2 (7-8-9) . it equals to zero

3) select a dog from Group 1 and one from Group 2 (again no littermate)

3*6=18

(12+0+18)/9C2=30/36=5/6

kapilhede17 · Answer

Well i find it easy to solve these relationship problems plainly by looking at the number of relationships.

Since 6 have exactly 1 littermate , 
what u really have are 3 relationships amongst 6 . 
1->2
3->4
5->6

Since 3 have exactly 2 relationships. 
what u really have are 3 relationships. 
7->8
8->9
7->9

Total of 6 relationships. 
Probability they will be relatied = number of relationships/ total number of ways u can pick 2 from 9 
= 6/ 9C2
=6/36=1/6

Hence Probablity that they are not related is = 1- 1/6 = 5/6

EvaJager · Answer

We have three pairs of dogs for the 6 with exactly one littermate, and one triplet, with each having exactly two littermates. 
So, in fact there are two types of dogs: those with one littermate - say A, and the others with two littermates - B.

Work with probabilities:
Choosing two dogs, we can have either one dog of type B or none (we cannot have two dogs both of type B).
The probability of choosing one dog of type B and one of type A is 3/9 * 6/8 * 2 = 1/2 (the factor of 2 for the two possibilities BA and AB).
The probability of choosing two dogs of type A which are not littermates is 6/9 * 4/8 = 1/3 (choose one A, then another A which isn't the previous one's littermate).
The required probability is 1/2 + 1/3 = 5/6.

Find the probability for the complementary event: choose AA or BB.
Probability of choosing two dogs of type A who are littermates is 6/9 * 1/8 = 1/12.
Probability of choosing two dogs of type B (who necessarily are littermates) is 3/9 * 2/8 = 1/12.
Again, we obtain 1 - (1/12 + 1/12) = 5/6.

Answer: C

Jp27 · Answer

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

If I name the dogs A thro I then then below are the pairs ->

AB
BA
CD
DC
EF
FE 
G H and I
H G and I
I G and H

To select a pair that is not a little mate then

I can select any of 9 dogs for the 1st slot....

then how to do i got about?
I could i have selected A in my 1st pick then i have remaining 7 to chose from (leaving B) = 9 * 7 / 2!

but If i select G 1st the I have 6 dogs left to chose from (leaving I and H) = 9 * 6 /2!

But this doesn't seem to work?

Bunuel please help

OA  C

actleader · Answer

How to understand 6 dogs have 1 littermate and 3 dogs have 2 littermates and the question is about dogs without littermates? :shock:

What does littermate in this case mean?

Jp27 · Answer

I just translated "littermate" to pair...
Although this is from kaplan I have seen more similar questions from Manhattan, which are quite strait forward for example:

there 8 ppl in a committee each have 1 sibling pair how many ways to select a 3 ppl who are NOT a sibling pair ->

8 * 6 * 4 / 3! = 32. 
this 1 is extension of this problem with more pairs hence more difficult.

actleader · Answer

it is harder because of added probability conditions. in the case with M's committees it's simply a combinatorial. so if i've understand you rightly - thare are aditionall 6+3 littermates along with 9 dogs or I'm blunting?

hiteshwd · Answer

Please help me with a fundamental understanding:

I solved this Q using the reverse technique i.e.:
Probability of not getting any littermates = prob of not getting (a. any of littermate doubles + b. any two of littermate triplets)
= (2/9)*(7/8) + (2/9)*(7/8) + (2/9)*(7/8) + (3/9)*(6/8)
= 7/12 + 1/4
= 5/6

So, the below method should work as well:
= (2C1*7C1)*3 / 9C2 + (3C1*6C1) / 9C2

But, it does not, I guess there is some problem with 9C2, because if I replace the same with 9*8, the equation becomes the same as above

I am not able to point out the fundamental flaw in 9C2 method as we use it quite often in similar Qs

For eg: 
A 10-member student leadership committee consists of 4 juniors and 6 seniors. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?

Sol: 
= 1- (4C4*6C2) / 10C6
= 13/14

Kindly help me to clear this fundamental flaw in my understanding. I have been banging my head over this for few hours:)

KarishmaB · Answer

Responding to a pm:

Finding littermates is much easier than finding non-littermates. There is much less confusion. Still, you can obviously work it out the other way around too.

Non littermates can be found in two ways:

1. You select one of the 3 littermates and one of the three pairs of 2 littermates.
3C1 * 6C1 = 18 ways

2. You select two of the three pairs of 2 littermates such that they belong to different litters.
You pick any one of the 6 dogs in 6 ways (say you picked B) and then you have 4 options (since you cannot pick A now). Say, you picked C.
There are 6*4 = 24 ways.
But wait, here, you have arranged the picked dogs. You have picked BC. In a different case, you would have picked C and then B. But both these are the same. So you need to divide 24 by 2 i.e. 12 ways.

Total number of ways of picking non-littermates = 18+12 = 30

Number of ways of picking 2 dogs = 9C2 = 36

Required Probability = 30/36 = 5/6

cledgard · Answer

By working directly with probabilities, I found non-littermates more directly and faster.
Since there are 6 dogs with exactly 1 littermate each, then we have three pairs of littermates: (1-2), (3-4), (5-6): group A
Since there are 3 dogs with exactly 2 littermates each, then we have one triple of littermates: (7-8-9): group B.
The probability of not littermates is: 1st selection a dog from group B over 9 dogs , 
2nd selection a dog from group A over the 8 remaining dogs :  3/9 x 6/8  
To this, we add the probability of 1st selection a dog from group A over 9 dogs, 2nd selection any of the other dods except for its mate over the 8 remaining dogs:  6/9 x 7/8  
3/9 x 6/8 + 6/9 x 7/8 = 5/6
 Choice C

BrentGMATPrepNow · Answer

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

cledgard · Answer

Exactly my point in the previous post, but you explained it with more detail.

EMPOWERgmatRichC · Answer

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:  C

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

NinetyFour · Answer

My thought process if it helps anyone:

out of 9 dogs we can pair them in 36 ways or  9C2

Probability we can pick a sibling + probability we don't pick a sibling = 1

To pick a sibling we have 3 pairs and one triple

So:

2C2 + 2C2 + 2C2 + 3C2 = 6

there is  \frac{1}{6}  chance to select a sibling

Hence  \frac{5}{6}  we do not select a sibling.

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