Problem Solving

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

Segment the problem:

2 Litter Mates : 3 dogs can have 2 litter mates each . Thus 3 ways

1 Litter Mate: 6 dogs if have 1 litter mate (LM) each then they will be in pair of two AB,BC,CA ie A is LM of B and B is LM of A (so 2 pair) . Thus 3 ways again

selecting either of the criteria , its becomes 3+3 so 6 ways to select 2 LMs .

Probability = 6 / 9C2 = 6/36

now choose ur ans .

BTW one note i dont think its a gmat question, as definition of LM is confusing ... expert may comment

nidhis.1408 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

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.

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