The probability that a point has max,min $x$,$y$ among $n$ points There are $n$ points randomly by uniform distribution distributed in the plane. We know that no two points have the same $(x ,y)$. What is the probability that a point has one of these properties?


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*it has maximum value of $x$ and maximum value of $y$ among $n$ points.

*it has maximum value of $x$ and minimum value of $y$ among $n$ points.

*it has minimum value of $x$ and maximum value of $y$ among $n$ points.

*it has minimum value of $x$ and maximum value of $y$ among $n$ points.


My solution:
First, I am going to answer a trivial question to clarify my solution. What is the probability that a point has minimum value of $x$ among $n$ points and what is the probability that this set contains a point with minimum value of $x$? Since there is no bound on $x, y$ and $n$ points are random, I assume exactly one point has the minimum value of $x$. The probability that every point has the minimum value of $x$ is $1/n$, So the probability that a set of $n$ points has a point with minimum value of $x$ is $\sum_{i=1}^n{1/n} = 1$. 
To answer the original question, There are $n$ different values of $x$ and $n$ different values of $y$ (I assume this is true, If not I should calculate the expected number of different $x$s and $y$s), the point must have a specific $(x,y)$, so the probability that a point has one of the above properties is $1/{n^2}$. So the probability that a set of points have one of the above properties is $4 * \sum_{i=1}^n{1/{n^2}} = 4/n$.
Is this a right answer to my question?
 A: The setup of this problem is as follows. There are $n$ pairs $(X_1, Y_1), \ldots, (X_n, Y_n)$, where each $X_i$ and each $Y_j$ are independently drawn from a Uniform distribution. (Note, this is actually a "distribution-free" problem so the fact that the random variables are uniformly distributed is inconsequential.) Suppose we randomly select the pair $(X_i, Y_i)$. It will be helpful if we let $A$ denote the event that $X_i$ is either the maximum or the minimum of the set of $X$ values: $A=X_i=\max\{X_i\} \cup X_i=\min\{X_i\}$. Similarly, let $B=Y_i=\max\{Y_i\} \cup Y_i=\min\{Y_i\}$. Mathematically, you're question can then be stated as $P(A \cup B)$. Using the inclusion-exclusion principle, we can rewrite this as
$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$
It is straightforward to show that $P(A)=P(B)=\frac 2 n$. Also, the events $A$ and $B$ are independent so we have that $P(A \cap B)=P(A)P(B)=\frac 4 {n^2}$. Therefore, $P(A \cup B)=\frac 4 n - \frac 4 {n^2}$. You're original answer was close but it didn't consider the fact that you were implicitly double-counting $P(A \cap B)$. However, if $n$ is large, then that term is very small.
