Expected number of isolated persons Suppose three persons are seated at random in a row having six seats. A person is said to be isolated if seats on his left and right sides are empty. Find the probability distribution of the number of isolated persons. Hence or otherwise find the expected number of isolated persons.
What I tried:-The question was asked in an examination. Due to time constraint, I did not ponder more and tried to do it by an unsophisticated way(By examining all the possible patterns).   
Case-I: Two(maximum) isolated persons. The possible patterns are-
_ $M_1$ _ $M_2$ _ $M_3$
$M_1$ _ $M_2$ _ $M_3$ _
In each of the two pattern the three persons can be arranged among themselves in $3!$ ways. So, total number of arrangements where two persons are isolated is $12$  
Case-II: One isolated person. The possible patterns are
_ $M_1$ _ $M_2$ $M_3$_
_ $M_1$ _ _$M_2$ $M_3$
$M_2$ _ $M_1$_ _ $M_3$
_ _ $M_1$_ $M_2$ $M_3$
$M_2$ _ _ $M_1$ _ $M_3$
$M_2$$M_3$ _ $M_1$ _ _
$M_2$$M_3$ _ _$M_1$ _
_$M_2$$M_3$ _ $M_1$ _   
Total number of arrangements is $8\times 3!=48$
Again, all possible cases is $\frac{6!}{3!}=120$
So, the number of arrangements where no one is isolated is $120-(48+12)=60$
The required probability distribution is
$P(X=0)=\frac{60}{120}=\frac{1}{2}$
$P(X=1)=\frac{48}{120}=\frac{2}{5}$
$P(X=2)=\frac{12}{120}=\frac{1}{10}$  
Now, $E(X)=0. \frac{1}{2}+1.\frac{2}{5}+2.\frac{1}{10}=\frac{3}{5}$  
Is my procedure correct?  
The questions also gives an opportunity to do the same work by some another method. What can be the other method?
 A: I find it odd that the question did not expect you to count the people sitting on the edges. I find the original question you had on the contest to be somewhat oddly phrased.
Moving on from that, yeah it is right; one minor comment: You don't need to care about who M1, M2, and M3 are, so you can just deal with them without caring about their internal orders. For instance, $\binom{6}{3}$ would give you the number of ways you can sit 3 people would be $\frac{6!}{3!3!} = 20$, and you found the probability distribution quite correctly I think.
The other way to use to find the expectation would be to use something called indicator functions, and use the linearity of expectation. You can imagine that the number of people sitting by themselves is just whether seats 2,3,4, and 5 have an isolated person on them or not. Now, the probability of each of them being isolated is 3/20 (only 3 cases left when you make a certain seat i to be isolated), and there are 4 of them, so that gives you your answer. 
Frankly though, I would not expect anyone to do this second method unless they took an actual introductory probability course. When there are 20 cases total, brute force really is not bad. 
