I recently turned in an assignment and had an error on it, or so I'm told, I'm not entirely convinced just yet. The problem was as follows:
$$F(x) =\begin{cases}1-\frac{16}{x^2}, & x\ge4 \\ 0, & x<4 \end{cases}$$
There were a number of different things I was to do with this problem. It wasn't until we were finding a few simple probabilities that the confusion set in.
It is my understanding that given $F$ is a distribution function of a random variable $X$, for an event concerning $X$, the probability of the event in terms of $F$ is as follows:
$$ \begin{array}{c|lcr} \text {Event}& \text{Probability} \\ \hline X \le a &F(a) \\ X \gt a &1-F(a) \\ X \lt a &F(a-)\\ X \ge a &1-F(a-)\\ X=a &F(a)-F(a-)\\ a\lt X\le b&F(b)-F(a)\\ a\lt X < b&F(b-)-F(a)\\ a\le X \le b&F(b)-F(a-)\\ a\le X < b&F(b-)-F(a-) \end{array} $$
where $F(a-)$ and $F(b-)$ represent the left hand limits. (From pg $146$ of Fundamentals of Probability with Stochastic Processes by Saeed Ghahramani)
I was asked to find $P(X \ge 6)$. What I did:
$$P(X \ge 6)=1-F(6-)=1-\frac{16}{36}=\frac{20}{36}=\frac59$$
What my professor had in his solutions (and also another student I spoke with agreed was correct) was:
$$P(X \ge 6)=1-P(X=6)$$ $$P(X=6)=1-\frac{16}{36}=\frac59$$
Thus, $$P(X \ge 6)=1-\frac59=\frac49$$
What am I missing here? Am I misunderstanding the chart?
The function $F(x)$ that you have described is clearly a cumulative distribution function. We have $\Pr(X\ge 6)=\Pr(X\gt 6)=1-F(6)=1-\left(1-\frac{16}{36}\right)=\frac{4}{9}$.
Our cdf is continuous. Thus there is no need to worry about the distinction between $\ge$ and $\gt$: For any $a$, the probability that the random variable is exactly equal to $a$ is $0$.
Remark: Note that $F(x)=\Pr(X\le x)$, and that $F(x)=1-\frac{16}{x^2}$ (for suitable $x$). The mitake in your calculation is a small slip. What you call $F(6-)$ should be $1-\frac{16}{36}$, not $\frac{16}{36}$.
By the way, although $F(6)$ is indeed $\frac{5}{9}$, it is not true that $\Pr(X=6)=\frac{5}{9}$. We have $\Pr(X\le 6)=\frac{5}{9}$, but $\Pr(X=6)=0$.