Find $P(X_{(7)}/X_{(10)} \leq a) $ where $X_{(n)} = \max(x_{1},\dots,x_{n}), x_{i} \sim \mathrm{Uniform}(0,1)$ Find $P(X_{(7)}/X_{(10)} \leq a) $ 
where $X_{(n)} = \max(x_{1},\dots,x_{n})$
$x_{i} \sim \operatorname{Uniform}(0,1)$
I found that $CDF: F_{n} (x) = [F(x)]^n  $
and $pdf: f_{n} (x) = n f(x) [F(x)]^{n-1} $
So I'm thinking of that
$$P(X_{(7)}/X_{(10)} \leq a) = P (X_{(7)} \leq a X_{(10)} \mid X_{(10)} = x)$$
$$=\frac {P(X_{(7)} \leq a X_{(10)}, X_{(10)} = x)}{P(X_{(10)} = x)}$$
$$=\frac {(xa)^7} {10 (x)^9} = \frac {a^7}{10x^2}$$
Does this all make any sense at all? Is this the right direction?
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Edit: How does this compare to $$P(x_2/x_3 \leq t)$$ ?
 A: The quantity $P(X_{(7)}/X_{(10)}\le a)$ does not depend on any number called $x$, so you should get something not depending on some number called $x$.  And just what $x$ is at that point is not clear.  It is correct that
$$
\Pr\left(\dfrac{X_{(7)}}{X_{(10)}}\le a\right) = \Pr(X_{(7)}\le aX_{(10)}).
$$
If you look at $\displaystyle\Pr\left(X_{(7)}\le aX_{(10)} \mid X_{(10)}=x\right)$, and find that it's some particular number, which let us call $g(x)$, then $\displaystyle\Pr\left(X_{(7)}\le aX_{(10)}\right)$ is actually the expected value of $g(X_{(10)})$.  That can be written as
$$
\Pr\left(X_{(7)}\le aX_{(10)}\right) = \mathbb E\left(\Pr\left(X_{(7)}\le aX_{(10)}\right) \mid X_{(10)}\right).
$$
(Google the term "law of total expectation".) That is actually a somewhat useful way to look at it if you can show that the conditional probability distribution of $X_{(1)},\ldots,X_{(9)}$ given the value of $X_{(10)}$ is actually the same as the distribution of the order statistics from a sample of size $9$ on the interval $(0,X_{(10)}$.  But it's even more convenient to look directly at
$$
\frac{X_{(1)}}{X_{(10)}},\ldots,\frac{X_{(9)}}{X_{(10)}}
$$
and show that that has the same distribution as the order statistics from a sample of size $9$ on the interval $(0,1)$.
Maybe I'll add more later.....
A: If $X\sim\operatorname{Uniform}(0,1)$, then the pdf of the sample maximum (in a sample of size $n$) is:  $n x^{n-1}$. Thus:


*

*if $x_{(n)}$ denotes the sample maximum in a sample of size $n$, and 

*if we have two independent samples, one of size 10, and the other of size 7, 


... then the joint pdf of $(x_{(10)}, x_{(7)})$, say $g(x_{10}, x_{7})$, is simply:

You seek $P(\frac{x_{(7)}}{x_{(10)}} \le a)$ which is:

where I am using the Prob function from the mathStatica package for Mathematica to do the pleasantries (I am an author of the former). All done.

As an aside, the pdf of your ratio is just the derivative of the cdf above wrt $a$, which yields:

Here is a plot of the theoretical pdf of $\frac{x_{(7)}}{x_{(10)}}$ (red dotted) and superimposed on top is a Monte Carlo simulation of the same ratio (always a great way to check for mistakes):

