Conditional probability with students seating There are 14 students and 9 of them are friends.
Students purchased tickets to movie and they got seats in a row of 14 seats.
8 friends got seats next to each other. What is the probability that remaining friend will get sit next to one of his friends?
My solution:


*

*Probability that 8 friends are seated at the beginning or at the end
of the row: $1/3$ 

*Probability that 8 friends are seated not in the beginning or end:
$2/3$

*Probability that remaining friend will get    seat next to his
friends: $1/3*1/6+2/3*2/6=5/18$
Any hints and solutions how to deal with this excersise would be appreciated
 A: I don't think that there is a $\frac{1}{3}$ chance that the octet will be at the beginning or end.  Wouldn't it be $\frac{2}{7}$?  Which would make the probability $\frac{2}{7}\frac{1}{6}+\frac{5}{7}\frac{2}{6}=\frac{2}{7}$
Alternately, you could think of the octet as a single entity.  There are $7!$ ways the group can be arranged.  Now lump the ninth friend in with the octet and there are now $6!\cdot2!$ ways this can be arranged.  $\frac{6!2!}{7!}=\frac{2}{7}$
A: I also got and answer of $\dfrac27$.
I believe this problem can be reduced to a simpler problem that you have probably solved before. I will start by treating the block of $8$ friends as one whole block. Since they are all assumed to be sitting together at the beginning, I can do this. That means that I am shrinking the problem from $14$ seats to $7$ seats total.
Now I am just trying to solve an analogous problem, I want to find the probability that "two people", the block of $8$ and the $1$ lonely friend sit together in a row of only $7$ seats. I think it would be simpler to approach this as a permutation. The total number of permutations would be $7!$ since anybody can sit in any seat. Now the numerator would be $(6 \times 2!)(5!)$ because in the set of $7$ seats there are only $6$ ways to sit pairs up, and there are $2!$ ways to arrange that pair. I multiply the the $5!$ because the other random people siting can sit where ever they want.
In conclusion $$\frac{(6 \times 2!)(5!)}{7!} = \frac27$$ when you simplify.
