Littermates- Probability

[email protected] wrote:Hi pareekbharat86,

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

The explanation in the test was incomprehensible. You have given a lucid explanation. Thanks!

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

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

Nice consistent answers, but I would like to offer a different answer - not on the list, as outlined below (followed by reasoning that both answers are correct depending on interpretation of the question).

Consider this scenario: the 3 dogs with exactly 2 littermates make a triangle, i.e. each corner of the triangle denotes one dog. The sides of the triangle each denote the sibling tie. So each dog in this triangle is a sibling of the other 2.

A - A
.\../
...A


Next, consider 3 separate pairs of siblings. All 6 have exactly one sibling/littermate.

B-B

B-B

B-B

This means that every dog is a littermates to either one or two others and the total is still 9 dogs.

Therefore the probability of any dogs not being littermates is zero.

ANSWER: P(not littermate) = 0.

What I mean is that there are no "only child" dogs; they are all littermates to some dog or another in the pack. Therefore it is certain that each and every dog chosen is a littermate. Therefore the probability of not being a littermate is always 0.

Therefore, to eliminate any perceived semantic ambiguity, may I propose that the question is rephrased as " If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates of each other". Now the previous solutions are the only correct interpretations.