good question

rishianand7 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?
1/6
2/9
5/6
7/9
8/9

Mitch has answered using the counting approach. Here's the 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
=[spoiler] 5/6[/spoiler]
= C

Cheers,
Brent