Conditioning within definition explanation I have a doubt on the meaning of a conditioning within a definition.
In a book I've found the following definition of upper tolerance limit:
$P(P(X<\bar X+kS|\bar X, S)>p)=1-\alpha$
where $X$ is a random variable, $\bar X$ is the sample mean of values taken from the distribution of $X$ and $S$ is their standard deviation. $p$ and $\alpha$ are numbers between 0 and 1.
The question is:
Why do we need the condition on $\bar X$ and $S$, and what does this conditioning mean in this context?
I.e. What's wrong with just writing the definition as
$P(P(X<\bar X+kS)>p)=1-\alpha$
 A: This is just an obfuscating way to express a simple idea: an upper tolerance limit is just an upper confidence limit for a percentile.
The random variable $X$ is supposed to have some definite (but unknown) distribution $F$.  The statistic $\bar X + k S$ (derived from a random sample from $F$) is intended to estimate the $p^\text{th}$ percentile of $X$, $F^{-1}(p)$.  That is, we hope that
$$F(\bar X + k S) = p.$$
Of course that won't be exactly true, because $\bar X + k S$ is random.  An upper tolerance limit is a procedure intended not to underestimate $F^{-1}(p)$.  The value of $\alpha$ is the chance you can tolerate of the procedure being wrong.  In other words, you want it to overestimate its target at least $1-\alpha$ of the time.  In many cases you can choose $k$ to assure this chance is exactly $1-\alpha$.  Thus,
$${\Pr}_F(F(\bar X + k S) \ge p) = 1-\alpha\tag{1}$$
is the defining criterion for a "$1-\alpha$ confidence upper tolerance limit of coverage $p$."  In English we could read it as

There is a $1-\alpha$ chance that the true percentile corresponding to the sample statistic $\bar X + k S$ will exceed $p$.

If you wanted to make expression $(1)$ look more complicated, you could unravel it using the definition of $F$; to wit,
$$F(z) = {\Pr}_F(X \le z)\tag{2}$$
for any real number $z$.  Fixing $z = \bar X + kS$ for the moment and plugging it into $(2)$ would give
$$F(\bar X + k S) = {\Pr}_F(X \le \bar X + k S).$$
That's a mighty ambiguous expression, though, because $F$ determines the distribution of both $X$ (thought of as an abstract random variable in $(2)$) as well as the distribution of $\bar X + k S$ (because that is determined by a random sample from $F$).  To make it clear we are talking in this context only of $X$ as the random variable, with $\bar X + k S$ being treated as a constant, we might write
$$F(\bar X + k S) = {\Pr}_F(X \le \bar X + k S\,|\, \bar X, S).$$
Plugging this into $(1)$ gives an expression like that in the book.  (It differs only in that I have been more careful in distinguishing $\ge$ and $\gt$, but that is of no matter.)

References
A standard book is Hahn & Meeker, Statistical Intervals, A Guide to Practitioners (John Wiley & Sons, 1991).  Here is its explanation:

The following characterizes a tolerance interval that one can claim contains a proportion $p$ of the population with $100(1-\alpha)\%$ confidence: "If one calculated such intervals from many independent groups of random samples, $100(1-\alpha)\%$ of the intervals would, in the long run, correctly include at least $100p\%$ of the population values..."

