How to calculate the probability of a point being inside a polygon Given that a point is in a polygon, I am assuming that this point is more likely to be on (or near) the Centroid of the polygon than it is likely to be on (or near) the edges of the polygon. Is that a correct assumption? If so, I need to prove that. And for doing so I think I need to know:  


*

*How can I calculate the probability of this point being on the centroid of the polygon?

*How can I calculate the probability of this point being on one the edges of the polygon?

 A: Picture a (regular) polygon with so many sides it is virtually indistinguishable from a circle.  The set of points that are closer to the centroid than to the any of the edges is, approximately, a circle of half the radius, hence roughly a quarter the area.  So your assumption appears to be incorrect.
A: Here's an example (too long for a comment) that might help us help you.
Suppose the polygon is the triangle with vertices $(0,0), (1,0), (0,1)$ and the point is chosen by selecting $x$ and $y$ uniformly each in $[0,1]$, throwing the answer away until you have $x+y \le 1$. Then all points in the triangle are equally likely, and any individual point has probability $0$. The probability is also $0$ that you're on an edge.
There might still be sense in asking for the probability of "near the centroid" or "near an edge" if you could say precisely what you mean by "near".
Edit - followup.
You can ask for the probability that a point is nearer the centroid than the boundary. For a circle that's just 25%. For a square it's less than 25%. For a very long very thin rectangle it's near $0$. I conjecture that for any convex figure it's always less than or equal to 50% - so you're more likely to be nearer the edge than the centroid. Less than or equal to 25% makes a stronger conjecture. 
A: Note: this is a wrong answer, as noted in the comments.

This is valid for a convex polygon. Take the centroid to be the origin and let the boundary of the  polygon be given in polar coordinates by
$$
r=f(\theta),\quad 0\le\theta\le2\,\pi.
$$
The inside of the curve 
$$
r\frac12f(\theta),\quad 0\le\theta\le2\,\pi.
$$
are the points closder to the centroid that to the border. Using the formula of the area in polar coordinates, the probability of beeing closer to the centroid turns out to be $1/4$.

Edit
Motivated by Barry Cipra's comment, I have done the calculations for a $(2\,L)\times 2$ rectangle, $L\ge1$.
 
The picture shows the part of the rectangle in the first quadrant. Blue points are closer to the origin than to the boundary; green points are closer to the boundary than to the origin. The boundary between the two regions is formed by two arcs of parabola of equation
$$
x^2+y^2=(1-x)^2,\quad\text{and}\quad x^2+y^2=(1-y)^2.
$$ 
The area of the blue region is
$$
\frac{4\,\sqrt2-5}{3}=0.218951\dots
$$
The probability of being closer to the center than to the boundary (assuming uniform distribution) is
$$
\frac{4\,\sqrt2-5}{3\,L}.
$$
A: Some thoughts on your question:
As a matter of convention, the term "random" in probability parlance has a different meaning than in usual speech.  A random variable (such as a point randomly selected within a polygon) is simply one that can take on any of an array of values, along with a probability distribution that describes how likely those values are.
Common examples are the flip of a coin, or the toss of a die, but it may be mildly surprising to know that loaded coins or dice still produce random variables.  Fair coins or dice produce random variables that we say are uniformly distributed, because each value is produced with equal probability: $1/2$ for each side of a coin, $1/6$ for each face of a die.  But a coin that comes up heads $55$ percent of the time is still random—it just doesn't produce a uniform distribution.  It isn't fair, in other words.
I mention this not to pick at your usage of the term "random", but to explain why the answers you've gotten in the comments might not be what you expected.  You said "random", but perhaps you meant "uniformly distributed".  (Or maybe not.  Read on.)
A second observation is that there is a basic distinction between what we call discrete random variables and continuous random variables.  Discrete random variables are the kind we've already mentioned: Like the flip of a coin or the toss of a die, they produce a range of values that you can enumerate, one by one.  We can therefore say, for each value, how likely that individual value is.
Continuous random variables, in contrast, take on (as their name implies) a continuous range of values.  The length of time it takes for the next bus to come is a popular example of a continuous random variable.  It could be $5$ minutes, it could be $3.14159$ minutes, it could be anything.  The fact that we rarely break down such times into units finer than a second might give the impression that the possible lengths of time can be enumerated (like the faces of a die), but that's an illusion: In principle, that time can be arbitrarily finely divided.
One consequence of such a time interval being a continuous random variable is that there are an infinite number of possible values (even though the time it takes is finite).  This in turn is a consequence of there being, paradoxically, an infinite number of numbers in any range of positive length.  That makes it difficult, at first glance, to assign a uniform distribution to those values.  How can you give each value an equal probability, when there are an infinite number of them?  It seems like each one would have to have zero probability.  But if each one has zero probability, how can they add up to one?
Such difficulties lead to a different way of specifying a probability distribution for continuous variables.  What we give instead is the probability "density".  For instance, suppose the probability density of the time till the next bus arrives is $0.1$ per minute over the interval from $0$ to $10$ minutes.  That means that if we want to know how likely it is that the bus will arrive in between $5$ and $8$ minutes from now, we take the length of that range ($8-5 = 3$ minutes) and multiply it by the density ($0.1$ per minute) to get the probability: $3 \times 0.1 = 0.3$.  In this case, the density is a function of time, and can be written, symbolically, as follows:
$$
f(t) = \begin{cases}
    \hfill 0.1 \hfill & 0 \leq t \leq 10 \\
    \hfill 0 \hfill & \text{otherwise}
\end{cases}
$$
This density can be used with any interval of time, no matter how small (provided it falls in the range from $0$ to $10$ minutes from now).  If we want to know how likely the bus will arrive in the interval of time between $2.71828$ and $3.14159$ minutes from now, we multiply $3.14159-2.71828 = 0.42331$ minutes by $0.1$ per minute, to get a probability of $0.042331$.
The limiting question might be, how likely is it that the bus will arrive exactly $5$ minutes from now.  And I mean, really really exactly: not $5.1$ minutes, not $4.99$ minutes, not even $5.000001$ minutes.  Exactly $5$ minutes.  The answer would be obtained by multiplying the length of that "interval" ($5-5 = 0$ minutes) by the density ($0.1$ per minute) to get an overall probability of $0$.  That is why the comments indicated that (under ordinary assumptions) the probability of a randomly selected point is zero.
Now, oddly enough, the probability being zero does not mean that it is impossible.  It just means that it is almost surely not $5$ minutes.  This distinction is hard to grasp at first, but it is a basic part of life when dealing with continuous random variables.
Thirdly (and lastly, or else I'll never eat lunch): Since some probability density (or distribution) functions—called PDFs for short—are uniform, it stands to reason that some are not.  If you plot the function $f(t)$ above, you'll see that it's a rectangle with height $0.1$ and length $10$.  As you become more comfortable with probability distributions, you will recognize that as the characteristic shape of the uniform distribution.
But there are cases where the uniform distribution is not appropriate.  We might decide, for instance, that a bus is more likely to arrive closer to $5$ minutes from now, compared to arriving in the next minute, or closer to $10$ minutes from now.  This might be represented by a probability distribution that starts out low, rises to a peak at $5$ minutes, and then drops down low again at $10$ minutes.  Something like a bell curve, for instance.  Or, it might be more likely for the bus to arrive very soon, and less likely later.
The point, though, is that all of these are probability distributions of random variables, and they all yield different answers to the question of what values are more likely.  So, when you ask the question of which point inside a polygon is more likely, it's necessary to describe the probability distribution.  That can be done by giving a function, like $f(t)$.  Or, it can be done by describing the process by which the point is selected.  One example might be as follows:

Pick two points $A$ and $B$ on the polygon's perimeter, according to a
  uniform distribution.  Find the midpoint $M$ of the line segment
  $\overline{AB}$.  Point $M$ is the selected point.

Then one might begin to address the question.  We might be able to say how the distribution varies within a regular $n$-sided polygon, and then progress to general statements about more irregular polygons.  But absent any kind of description of how the point is selected, it is not possible to give the question (of what points are more likely) any kind of meaningful answer.
Hope this helps you (and anyone else who reads this).
