Density of primes in intervals less than primorial numbers Looking at the interval of the natural numbers $ [1, p_{n}$#$] $;
$\frac{1}{2}$ of the elements of this set will be even, and $\frac{1}{2}$ will be odd.
$\frac{1}{3}$ of the elements of this set will be multiples of 3, and $\frac{2}{3}$ will not ($:n>1$) ...  $\frac{1}{p_{n}}$ of the elements will be multiples of $p_{n}$, and $\frac{p_{n}-1}{p_{n}}$ will not.
The proportion of elements of this set that are not multiples of any prime up to and including $p_{n}$ will be: $\frac{1}{2} \times \frac{2}{3} \times ... \times \frac{p_{n}-1}{p_{n}}$ ($:n>2$), assuming independency of these proportions (true?). This is the same as the expression $ 1 \times \frac{2}{2} \times \frac{4}{3} \times ... \times \frac{p_{n}-1}{p_{n-1}} \times \frac{1}{p_{n}} $ (: $n>4$), which because $ p_{n}-1 \geq p_{n-1} $; must be larger than $\frac{1}{p_{n}} $. 

i.e. At least $\frac{1}{p_{n}} $ of elements of the set $ [1, p_{n}$#$] $ are not multiples of any prime numbers up to and including $p_{n}$.

So I ask, under what conditions will this proportion hold, that atleast $\frac{1}{p_{n}}$ elements are not divisble by prime numbers upto $p_{n}$; within an interval that is contained by $ [1, p_{n}$#$] $?
At first I thought if the size of the interval was a multiple of $p_{n}$ which is less than or equal to $p_{n}$#, this would be sufficient, but I know this isn't true. 
However it does seem true if the largest term of this interval is a multiple of $p_{n}$ (must be $\leq p_{n}$#), as long as the size of the set is also a multiple of $p_{n}$ (can be different but must also be $\leq p_{n}$#).
Is this just a coincidence or is there a reason behind this?
Examples:
1. $p_{2}=3$ 
The primorial number $p_{2}$# $ = 2\times 3 = 6$
In the interval [1,6], half of the members are even numbers, a half of the members are odd, a third of the members are multiples of 3, and two thirds of the members are not multiples of 3.
Consequently, at least a third of the members are not divisible by 2 or 3. 
Similarly in the intervals [1,3] and [4,6], at least a third of their members are not divisible by 2 or 3. Is this just coincidental or is their a reason behind it? Can you derive the conditions of an interval having at least a third of its members not divisble by 2 or 3, with respect to the interval [1,6]?
I ask because in the interval [2,4], the proportion of members that are not divisible by 2 or 3, is not at least a third. 
2. $p_{3}=5$
The primorial number $p_{3}$# $ = 2\times 3 \times 5= 30$
In the interval [1,30], half of the members are even numbers, a half of the members are odd, a third of the members are multiples of 3, and two thirds of the members are not multiples of 3, a fifth of the members are multiples of 5, and four fifths of the members are not multiples of 5.
Consequently, at least a fifth of the members are not divisible by 2 or 3 or 5. 
Similarly in the intervals [1,5], [6,10], [11,15], [16,20], [21,25], [26,30], [1,10], [11,20], [21,30], [6,15], [16,25], [1,15], [16,30], [1,20], [11,30], at least a fifth of their members are not divisible by 2 or 3 or 5. Is the similarity in proportions, between these and the interval [1,30] just coincidental or is their a reason behind it?
(In the interval [24,28], the proportion of members that are not divisible by 2 or 3 or 5, is not at least a fifth.)
 A: I encourage you to buy Hardy and Wright, An Introduction to the Theory of Numbers. I have the fifth edition, a sixth edition came out recently with added material...
Anyway, let us write the specific primorial $P$ as
$$  P = \prod_{p \leq x} p.  $$
The largest prime dividing $P$ is the largest prime no larger than $x,$ and is very similar in size to $x$ itself.
Now, You are asking about $\phi(P),$ where $\phi$ refers to Euler's Totient function. Your first observation is that
$$  \frac{\phi(P)}{P} =   \prod_{p \leq x} \left( 1 - \frac{1}{p} \right).  $$
For the entire "interval," you would like to know that this is larger than $\frac{1}{x}.$ That is, you would like to say that $$ x \, \frac{\phi(P)}{P} > 1.   $$
Well, Theorem 429 in H+W, Merten's Theorem, is that 
$$  \frac{\phi(P)}{P} =   \prod_{p \leq x} \left( 1 - \frac{1}{p} \right) \sim \frac{e^{-\gamma}}{\log x}.  $$
As a result, your
$$ x \, \frac{\phi(P)}{P} =  x \, \prod_{p \leq x} \left( 1 - \frac{1}{p} \right) \sim \frac{e^{-\gamma} x}{\log x},  $$ which grows without bound, and certainly stays larger than $1.$ Oh, here $e\approx 2.718281828$ is one of Euler's constants, and the logarithm is base $e.$ 
For large enough $x,$ meaning the largest prime $p \leq x$ is also large, one would expect your subintervals, if long enough, to also have a high density of numbers not divisible by any prime up to $x.$
Meanwhile, $\gamma \approx 0.5772156649$ is also called Euler's constant.
