In any $n$ consecutive odd integers, there are $\lfloor n/3\rfloor$ disjoint triples of $3$ consecutive odd integers, so at least that many
of the integers are divisible by $3$. Therefore (if all the integers are greater than $3$)
there are at least $\lfloor n/3 \rfloor$ composites and at most $n - \lfloor n/3 \rfloor = \lceil 2n/3 \rceil$ primes.
EDIT: For computations involving $\lfloor \ldots \rfloor$ and $\lceil \ldots\rceil$, it's often best to "unwind" the definition. $x = \lfloor n/3 \rfloor$ means that $x$ is an integer with $x \le n/3 < x + 1$. Then
$n - \lfloor n/3 \rfloor = n - x$ is an integer with
$n - x - 1 < 2n/3 \le n - x$. And that's the definition of $\lceil 2n/3 \rceil$.