CMIIh

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.

Nathanael Webber

Skilled2020-11-10Added 117 answers

Step 1

We can pick 4 people from the 8 people for the first room in $\text{}}_{8}{C}_{4$ ways

When 4 people are picked then we can pick rest 4 people in $\text{}}_{4}{C}_{4$ 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 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 $\text{}}_{6}{C}_{2$ ways

And the rest 4 people can be selected for the second room in $\text{}}_{4}{C}_{4$ ways

So the total number of ways is $=\left({}_{6}{C}_{2}\right)\left({}_{4}{C}_{4}\right)=15\left(1\right)=15$

For the second case: We have to pick 4 people from the 6 people for the first room

We can do that in $\text{}}_{6}{C}_{4$ ways

And we pick rest 2 people in $\text{}}_{2}{C}_{2$ 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 different ways.

So the required probability is $=\frac{30}{70}=\frac{3}{7}$

Hence, $\frac{3}{7}$

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