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 e that both selected dogs are not littermates? Select
A) 1 / 6
B) 2 / 9
C) 5 / 6
D) 7 / 9
E) 8 / 9
<gadget id="calculator">5 / 6</gadget>
<output>5/6 = around 0.833333</output>

<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 e that both selected dogs are not littermates?
divide(const_5, 6)
"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 can not 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 e 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 is n ' 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"