Where do I go wrong with my calculation?
Let's assume $\xi_{1}\sim Uni[0,40]$, $\xi_{2}\sim Uni[0,100]$ and $\eta$ is a Bernoulli variable with $p=0.5$ Let's assume $\xi_{1}$, $\xi_{2}$ and $\eta$ are independent. What is the following probability and where do I go wrong with my calculation?
$$\begin{align*} \mathbf{P}\left(\eta\cdot\xi_{1}+\left(1-\eta\right)\cdot\xi_{2}\geq20\right)= & \mathbf{P}\left(\eta\cdot\xi_{1}+\xi_{2}-\eta\cdot\xi_{2}\geq20\right)=\\ = & \mathbf{P}\left(\eta\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\right)=\ldots \end{align*}$$
Using the law of total probability
$$\begin{align*} \ldots & =\mathbf{P}\left(\eta\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\mid\eta=0\right)\mathbf{P}\left(\eta=0\right)+\mathbf{P}\left(\eta\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\mid\eta=1\right)\mathbf{P}\left(\eta=1\right)=\ldots \end{align*}$$
Here
$$ \begin{align*} \mathbf{P}\left(\eta\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\mid\eta=0\right) & =\frac{\mathbf{P}\left(\left\{ \eta\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\right\} \cap\left\{ \eta=0\right\} \right)}{\mathbf{P}\left(\eta=0\right)}=\frac{\mathbf{P}\left(0\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\right)}{1/2}=\\ & =2\mathbf{P}\left(\xi_{2}\geq20\right) \end{align*} $$
and similarly
$$ \begin{align*} \mathbf{P}\left(\eta\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\mid\eta=1\right) & =\frac{\mathbf{P}\left(\left\{ \eta\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\right\} \cap\left\{ \eta=1\right\} \right)}{\mathbf{P}\left(\eta=1\right)}=\frac{\mathbf{P}\left(1\left(\xi_{1}-\xi_{2}\right)+\xi_{2}\geq20\right)}{1/2}=\\ & =2\mathbf{P}\left(\xi_{1}\geq20\right) \end{align*}$$
So
$$\begin{align*} \ldots & =2\mathbf{P}\left(\xi_{2}\geq20\right)\frac{1}{2}+2\mathbf{P}\left(\xi_{1}\geq20\right)\frac{1}{2}=\\ & =\int_{20}^{100}\frac{1}{100}dy+\int_{20}^{40}\frac{1}{40}dx=\left[\frac{y}{100}\right]_{20}^{100}+\left[\frac{x}{40}\right]_{20}^{40}=\\ & =\left(1-\frac{1}{5}\right)+\left(1-\frac{1}{2}\right)>1 \end{align*}$$
See my comment for your mistake.
Further your attempt is cumbersome.
It is better to go more directly for:$$\mathbf{P}\left(\eta\cdot\xi_{1}+\left(1-\eta\right)\cdot\xi_{2}\geq20\right)=$$$$ \mathbf{P}\left(\eta\cdot\xi_{1}+\xi_{2}-\eta\cdot\xi_{2}\geq20\mid\eta=0\right)P(\eta=0)+\mathbf{P}\left(\eta\cdot\xi_{1}+\xi_{2}-\eta\cdot\xi_{2}\geq20\mid\eta=1\right)P(\eta=1)=$$$$\mathbf{P}\left(\xi_{2}\geq20\mid\eta=0\right)\frac12+\mathbf{P}\left(\xi_{1}\geq20\mid\eta=1\right)\frac12=$$$$\mathbf{P}\left(\xi_{2}\geq20\right)\frac12+\mathbf{P}\left(\xi_{1}\geq20\right)\frac12$$ The last equality rests on the fact that $\eta$ and $\xi_i$ are independent for $i=1,2$.