Given a $4\times 4$ Matrix having $16$ points, what is the probability of making a triangle? 
Given a $4\times 4$ Matrix having $16$ points, what is the probability of making a triangle from these points?

My Approach:
$4\times 4$ matrix has $16$ points. So, I can choose 3 point of a triangle from these $16$ points= $16\choose 3$
Is this correct?
 A: We are presumably choosing $3$ points from the $16$, with all choices equally likely. There are $\binom{16}{3}$ such choices.  To find the probability of making a triangle, we can count the number $F$ of favourable choices, where the $3$ points chosen form a triangle. Then the required probability is $\frac{F}{\binom{16}{3}}$.
We can either find $F$ directly, or count the number $B$ of bad choices, where the points chosen do not form a triangle, because they lie on a line. Then $F=\binom{16}{3}-B$.
How many ways could we choose badly? To do the counting, it is very useful to draw a careful picture of the $4\times 4$ grid, in order to spot the lines that have $3$ or more points of our grid. Think of the grid points as all points with coordinates $(i,j)$, where $i$ and $j$ are integers with $0\le i\le 3$ and $0\le j\le 3$.
We could choose badly by choosing $3$ points on the top row. There are $\binom{4}{3}$ ways to do this. Similarly, we could choose $3$ points on the second row, the third, the fourth, the left column, the next column, and so on.  That gives (so far) $8\cdot\binom{4}{3}$ bad choices.
We could also choose badly by choosing $3$ points from the diagonal joining $(0,0)$ to $(3,3)$, or from the other long diagonal that joins $(3,0)$ to $(0,3)$. That's $2\cdot \binom{4}{3}$ more bad choices.
That takes care of the lines that have $4$ points of our grid. But there are also lines that have $3$ points of our grid. One example is the line that goes through $(1,0)$, $(2,1)$, and $(3,2)$. By symmetry there are $3$ other such lines. That gives us $4$ more bad choices of $3$ points.
Any others? A careful look at our diagram shows there are no more lines that contain $3$ or more grid points. We end up with $B=44$.
