A group of $200$ persons consisting of $100$ men and $100$ women is randomly divided into $100$ pairs of $2$ each.Find the maximum chance that at most $30$ of these pairs will consist of a man and a woman.
Solution : Let m1,m2,m3.....m100 be the men and w1,w2,w3,....w100 be the women.
Let X1 be the random variable such that
X1= 1 if m1 is paired with some wj
=0 if m1 is paired with some mj
Similar is true for X2,X3,...,X100
In the way we construct the random variables, obviously they are dependent and if we define a random variable as ,
X=X1+X2+X3+....+X100, then the spectrum of X=$0$ ,$1$ ,$2$ ,$3$ ,$4$,...,$100$
For the m1, probability that he is paired with a woman is $100/199$
Probability that the man is paired with another man is $1-100/199 = 99/199$
THE SAME IS TRUE FOR ALL THE MEN FROM $2$ TO $100$
Now the problem turns out be $P(X<=30)$,
Here we can easily apply one-sided Tchebycheff's Inequality and get the result...
My question lies somewhere else,
The way the solution to the problem defined the variable $X$ , $X$ actually now defined the event that $X=k$, the people are divided such that there are $k$ pairs containing men and women.
Now i can't understand the way the m1 getting a woman is mutually exclusive to the situation when m2,m3,..m100 gets a woman.
The way Xi has been defined , it is quite logical for the ith man to have the possibility of having all the rest of woman, but but i feel that is so when a sort of CHOOSING WITH REPLACEMENT IS THERE, but when we are dividing into groups there is a sort of conditional probability that comes into play,that when m1 takes up a woman, m2,m3,.. can't possibly take up that woman. However if m2 takes up that woman, then m1,m3,..shall not be able to take up that woman...
Please can anyone explain me the situation in a clear cut solid way. Thanks in advanced.