Question

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. if two individuals are selected from the room at random, what is the probability that those two individuals are not siblings?

Answer 1

16/21, or approximately 76.19% change of not being siblings.  First, we need to determine the siblings. Since 4 people have exactly 1 sibling, that could only be achieved by having 2 pairs of siblings. Then the remaining 3 people are all siblings of each other. To illustrate, I'll use letters to represent each person. All people with the same letter are siblings to everyone else with that letter. So we have A,A,B,B,C,C,C in the room which meets the description of siblings. Since we're picking 2 people out of 7, there is 7!/(2!5!) = 5040/(2*120) = 5040/240 = 21 different ways of selecting those 2 people. Now it's easier to list those combinations that are siblings and then the number of pairs that are not is simply the complement. So let's do that. There's exactly 1 pair that are "A" siblings, and exactly 1 pair that are "B" siblings, and 3 pairs that are "C" siblings (3!/(2!1!) = 6/(2*1) = 6/2 = 3). So there are 1+1+3 = 5 possible pairs out of 21 that are siblings. That means that there's 21-5 = 16 possible pairs that are not siblings. Therefore the desired probability is 16/21, or approximately 76.19% chance of the pair not being siblings.