Xrfgtjtk

I would like to know how to compute the following conditional probability.
Let $X|theta sim mathcal{U}(0,theta)$ be a uniform random variable.
Given a fixed $theta = theta_0 > 0$, one can easily compute the conditional probability
$$
P(X le 1 | theta = theta_0) = int_0^{min{1,theta_0}} frac{1}{theta_0}dx.
$$

But I am not sure how one can compute the conditional probability of
$X$ given $theta < 1.5$ or $theta ne 1.5$, i.e., $$ P(X le 1 |
theta < 1.5) = ?, qquad P(X le 1 | theta ne 1.5) = ? $$

However, given a probability distribution on $theta$, I can compute
the above probabilities.
For example, assuming $theta sim mathcal{U}(0,2)$,
since
begin{align*}
P(Xle 1, theta le 1.5) &= int_0^{1.5} int_0^{min{1, z}} f(x|z)f_theta(z) dx dz
= int_0^{1.5} int_0^{min{1, z}} frac{1}{2z} dx dz \
&= int_0^{1.5} frac{min{1,z}}{2z} dz
= int_0^{1} 0.5 dz + int_1^{1.5} frac{1}{2z} dz \
&= 0.5(1 + ln 1.5),
end{align*}
one can compute
begin{align*}
P(X le 1 | theta < 1.5) &= frac{P(X le 1, theta < 1.5)}{P(theta < 1.5)}
= frac{0.5(1+ln1.5)}{0.75} = frac{2}{3}(1+ln 1.5).
end{align*}
Similarly, in order to compute $P(X le 1 | theta ne 1.5)$,
I considered
$$
P(X le 1, theta ne 1.5) = P(X le 1, theta < 1.5) + P(Xle 1, theta > 1.5).
$$
Since
$$
P(X le 1, theta > 1.5) = int_{1.5}^2 int_0^{min{1,z}} frac{1}{2z}dxdz = 0.5 ln frac{2}{1.5},
$$
we have
$$
P(X le 1, theta ne 1.5) = 0.5(1 + ln 2).
$$
It then follows from $P(theta ne 1.5) =1$ that
$$
P(X le 1 | theta ne 1.5) = 0.5(1+ln 2).
$$

Does it even make sense writing the following expressions without introducing a probability on $theta$?
$$ P(X le 1 | theta ne theta_0)? quad P(X le 1 | theta < theta_0)?
quad P(X le 1 | theta > theta_0)? $$
If it makes sense, I would like to know how to compute such probabilities.

Any comments/suggestions/answers will be very appreciated. Thanks.