Which is the best way to solve the below problem.
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?
Ans: 5/21, 3/7 ,4/7 ,5/7 ,16/21.
Solution 1: 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.
However , i found this technique is kind of specific to each problem and it is difficult to calculate individual strategy to each probability problem when u have like 1 or 2 mins.
My general strategy is always to take individual probability like i would go about the following way to calculate 2 siblings
4/7 * 1/6 + 3/7 * 2/6 = 5/21.
But even the second methods needs an exhaustive probability count for each and every scenario, and i tend to skip some scenarios resulting in wrong answers.
Is there a better approach to handle such probability problems ??
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?
Ans: 5/21, 3/7 ,4/7 ,5/7 ,16/21.
Solution 1: 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.
However , i found this technique is kind of specific to each problem and it is difficult to calculate individual strategy to each probability problem when u have like 1 or 2 mins.
My general strategy is always to take individual probability like i would go about the following way to calculate 2 siblings
4/7 * 1/6 + 3/7 * 2/6 = 5/21.
But even the second methods needs an exhaustive probability count for each and every scenario, and i tend to skip some scenarios resulting in wrong answers.
Is there a better approach to handle such probability problems ??
