$n$ men and $m$ women are standing in a line (randomly).
Find the expectancy of the number of men that stand beside a women (at least one side - left or right)
Harder question: Now solve it, but they stand in a circle and not a line.
I wanted to use an indicator:
$X = $ number of men standing beside at least one women.
$X_i = 1 ~~~\text{if standing besides a women} \\ ~~~~~~~~~~0~~~ \text{else}$
And so: $E[X] =E[\sum_{i=1}^nX_i] =\sum_{i=1}^nE[X_i]$
the problem is that I am having a time computing what is the probability a random men will stand next to a women (left or right or both)
+ How should I approach the harder question?
Thank you very much!
I do not really understand the nature of your problem with the first question.
It is a good strategy to go for indicators here.
In what you call the "harder question" we have: $$\mathbb EX_1=P(X_1=1)=1-P(\text{man 1 has male neighbors})=$$$$1-\frac{\binom{n-1}2\binom{m}0}{\binom{n+m-1}2}=1-\frac{n-1}{n+m-1}\frac{n-2}{n+m-2}$$
Moreover by symmetry we have $\mathbb EX_i=\mathbb EX_1$ for every $i\in\{1,\dots,n\}$ so that:$$\mathbb EX=n\mathbb EX_1=n\left(1-\frac{n-1}{n+m-1}\frac{n-2}{n+m-2}\right)$$
addendum:
To find $\mathbb EX_1=P(X_1=1)$ in the first problem is a bit harder.
If $E$ denotes the event then man 1 is at utmost left or utmost right position then we must split up:
We must split up in:$$P(X_1=1)=P(X_1=1\mid E)P(E)+P(X_1=1\mid E^{\complement})P(E^{\complement})$$
Here $P(E)=\frac2{n+m}$ and $P(X_1=1\mid E)=\frac{m}{n+m-1}$.
For $P(X_1=1\mid E^{\complement})$ we find what we found for $P(X_1=1)$ in the first problem.