Exponential random variable X with a uniform random variable as its parameter $$X\ \sim Exp(U) ~ and\ U\ \sim  U(0,1) $$
The question asked for the value of $ P(X\geqslant 1)$
I saw the solution and it went like this:
$$P(X\geqslant 1) = E[P(X\geqslant 1)|U] = E[e^{-u}] = \int_{0}^{1}e^{-x}dx  $$
Could someone explain the solution to me?
More specifically how did the solution derive each step?

You are given that the conditional distribution of $X$ given $U=u$ is $\text{Exp}(u)$ (presumably $u$ is the rate parameter), where $U$ itself is uniformly distributed on $(0,1)$.
The solution is an application of the law of total expectation, or equivalently, the law of total probability.
Suppose $I(X\ge 1)$ is an indicator variable that equals $1$ if $X\ge 1$ and equals $0$ otherwise. And let the density of $U$ be $f_U$. Then,
\begin{align}
P(X\ge 1)&=E\left[I(X\ge 1)\right]
\\&=E\left[E\left[I(X\ge 1\mid U\right]\right]
\\&=E\left[P(X\ge 1\mid U)\right]
\\&=E\left[e^{-U}\right]
\\&=\int e^{-x}f_U(x)\,dx
\\&=\int_0^1 e^{-x}\,dx
\end{align}
In the penultimate step, law of the unconscious statistician was used to find the expectation.
