CMIIh

Answered question

2020-11-09

Consider a group of 8 people. Among them, there is one pair of twins. These 8 people are taken into two different rooms, Room A and Room B, with four people to each room. If all groups of four people are equally likely, find the probability that the twins will be sent into the same room.

Answer & Explanation

Nathanael Webber

Skilled2020-11-10Added 117 answers

Step 1
We can pick 4 people from the 8 people for the first room in  ways
When 4 people are picked then we can pick rest 4 people in  ways
So total number of ways
So total number of ways to arrange the people $=70$
Step 2
If the twins are placed in the same room, two scenarios are possible.
2) Twins are in the second room
For the first case: We have to pick 2 more people for the first room from the 6 other people. We can do that in  ways
And the rest 4 people can be selected for the second room in  ways
So the total number of ways is
For the second case: We have to pick 4 people from the 6 people for the first room
We can do that in  ways
And we pick rest 2 people in  ways for the second room
So total number of ways
So for both cases total number of ways $=15+15=30$
So the twin can be together in 30 different ways.
So the required probability is $=\frac{30}{70}=\frac{3}{7}$
Hence,  $\frac{3}{7}$

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