***I don't understand the part in bold. How do you get 21 different ways to choose two people?
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?
a) 5/21
b) 3/7
c) 4/7
d) 5/7
e) 16/21
Answer: We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.
Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.
Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 – 5/21 = 16/21.
The correct answer is E.
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?
a) 5/21
b) 3/7
c) 4/7
d) 5/7
e) 16/21
Answer: We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.
Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.
Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 – 5/21 = 16/21.
The correct answer is E.
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