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.
Best Answer
As you say, the chance the first pair is a man and a woman is $\frac {100}{199}$. This means the expected number of man-woman pairs is $\frac {10^4}{199}\approx 50.25$ We expect that the chance of $30$ is quite small. If we draw with replacement the chance becomes $\frac 12$ and the correlations disappear. We can say the mean will be $50$ and the standard deviation $\sqrt{100 \cdot \frac 1{2^2}}=5$. We are four standard deviations away from the mean, so the chance will be very small. I suspect you are expected to use a z-score table at four standard deviations and find the chance of being outside that is about $3 \cdot 10^{-5}$
If we want to be more exact, we can compute the chance of exactly $k$ matches. $k$ must be even. There are $200!$ ways to put the people in order. When we make each pair there are $2^{100}$ ways to reverse the people in a pair and $100!$ ways to order the pairs, so the number of pairings is $\frac {200!}{2^{100}100!}$. To get $k$ man-woman matches we can select $k$ men and $k$ women in $100 \choose k$ ways each. We can pair them up in $k!$ ways, then pair the remaining men in $\frac {(100-k)!}{2^{k/2}(50-\frac k2)!}$ and the women similarly, so the chance of $k$ mixed pairs is $$\frac{\left({100 \choose k}\frac {(100-k)!}{2^{k/2}(50-\frac k2)!}\right)^2}{\frac {200!}{2^{100}100!}}$$