Expected number of women sitting next to at least one man? There are $10$ seats, $5$ men and $5$ women who randomly occupy these seats. I have to calculate the expected number of women sitting next to at least one man.
My attempt: I defined a random variable $X_i$ which takes value $0$ if seat number $i$ is occupied by a woman and all adjacent seats are occupied by women, and $1$ otherwise. 
For a corner seat, my expected value of $X_1$ turned out to be $(5/9)$
Because probability that one man occupies seat $2$ given seat $1$ is occupied by a woman is $5/9$.
For the non-corner seats, I calculated the expected value of $X_2$ to be $(15/18)$
Because the probability that seat $2$ is occupied by a woman and at least one of seat $1$ and $3$ is occupied by a man is $1-($prob all three occupied by women$)$
I calculated the final expected value of $X$ by adding all the expected values of $X_i$ (for $i=1,2...10$) and obtained $(70/9)$. This is obviously wrong, since the number of total women cannot be less than the expected number of women sitting next to at least one man. I'd like to know where I have made the error
 A: You had a good idea, but it's not quite right. You're counting not only seats occupied by women not surrounded by women but also all seats occupied by men. (You also didn't execute the idea correctly – for corner seats you calculated a conditional probability, and for interior seats I'm not sure exactly what you calculated.)
Here are three correct ways to solve the problem:
1) Perhaps most similar to what you tried to do, we can count the number of seats occupied by a woman and adjacent to at least one man. This is the probability for the seat to be occupied by a woman minus the probability for the seat to be occupied and surrounded by women. For a corner seat, this is
$$
\frac12-\frac{\binom83}{\binom{10}5}=\frac5{18}\;.
$$
For an interior seat, it's
$$
\frac12-\frac{\binom72}{\binom{10}5}=\frac5{12}\;.
$$
The total is
$$
2\cdot\frac5{18}+8\cdot\frac5{12}=\frac{35}9\;.
$$
2) Similarly, you can count the number of seats occupied and surrounded by women, and subtract that from the total number of women. For a corner seat, this is
$$
\frac{\binom83}{\binom{10}5}=\frac29\;,
$$
for an interior seat, it's
$$
\frac{\binom72}{\binom{10}5}=\frac1{12}\;,
$$
and so the expected value is again
$$
5-\left(2\cdot\frac29+8\cdot\frac1{12}\right)=\frac{35}9\;.
$$
3) Alternatively, you could focus on the women instead of the seats. Each woman has a probability of $\frac2{10}$ of being in a corner seat, and then she has a probability of $\frac59$ of sitting next to a man, and a probability of $\frac8{10}$ of being in an interior seat, and then she has a probability of
$$
1-\frac{\binom72}{\binom94}=\frac56
$$
of sitting next to a man, again for a total of
$$
5\left(\frac2{10}\cdot\frac59+\frac8{10}\cdot\frac56\right)=\frac{35}9\;.
$$
