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 Q that both selected dogs are NOT littermates?
Select: A) 1/6 B) 2/9 C) 5/6 D) 7/9 E) 8/9
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 Q= 1/12.   Again, we obtain 1 - (1/12 + 1/12) = 5/6.   Answer: 5/6
<result>C</result>
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 Q that both selected dogs are NOT littermates?