# 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.

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.
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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$=\left({}_{8}{C}_{4}\right)\left({}_{4}{C}_{4}\right)=70\left(1\right)=70$
So total number of ways to arrange the people $=70$
Step 2
If the twins are to be sent in the same room then two cases can happen
1) Twins are in the first room
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 total number of ways $=\left({}_{6}{C}_{2}\right)\left({}_{4}{C}_{4}\right)=15\left(1\right)=15$
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$=\left({}_{6}{C}_{4}\right)\left({}_{2}{C}_{2}\right)=15\left(1\right)$
So for both cases total number of ways $=15+15=30$
So the twin can be together in 30 ways
So the required probability is $=\frac{30}{70}=\frac{3}{7}$
Answer: $\frac{3}{7}$