probability that minimum value in set A is larger than maximum value in set B There are two sets, called $A$ and $B$ which have $K$ number of elements, respectively.
Each element in both group follows one of two exponential distribution.
One is $f_X(x) = \alpha e^{-\alpha x}$ (easily called alpha exponential), the other is $f_X(x) = \beta e^{-\beta x}$ (beta exponential), where $\beta \ll \alpha$ and $\alpha,\beta > 0$.
Each element in set $A$ follows alpha exponential with probability 0.2 and follow beta exponential with probability 0.8.
Also, each element in set $B$ follow alpha exponential with probability 0.8 and follow beta exponential with probability 0.2.
I want to find the probability that the smallest elements in set $A$ is larger than the largest value in set $B$.
How can I find the probability that the smallest elements in group A is larger than the largest value in group $B$?
 A: Let $A_1,\ldots, A_K$ be i.i.d. random variables with density
$$
f_A(t) = \frac15\alpha e^{-\alpha t} + \frac45\beta e^{-\beta t}
$$
and $B_1,\ldots, B_K$ be i.i.d. random variables, independent of the $A_i$, with density
$$
f_B(t) = \frac45\alpha e^{-\alpha t} + \frac15\beta e^{-\beta t}.
$$
The corresponding distribution functions are
$$
F_A(t) = 1 - \left(\frac15\alpha e^{-\alpha t} + \frac45\beta e^{-\beta t}\right),\quad F_B(t) = 1-\left(\frac45\alpha e^{-\alpha t} + \frac15\beta e^{-\beta t}\right).
$$
Let $\mathcal A=\min_i A_i$ and $\mathcal B=\max_i B_i$. Then
\begin{align}
\mathbb P(\mathcal A>t) &= \mathbb P\left(\bigcap_{i=1}^K \{A_i > t\} \right)\\
&= \prod_{i=1}^K \mathbb P(A_i>t)\\
&= \left(\frac15\alpha e^{-\alpha t} + \frac45\beta e^{-\beta t}\right)^K,
\end{align}
so the distribution function of $\mathcal A$ is
$$
F_{\mathcal A}(t) = 1 - \left(\frac15\alpha e^{-\alpha t} + \frac45\beta e^{-\beta t}\right)^K.
$$
By a similar computation we find that
$$
F_{\mathcal B}(t) = \left(1-\left(\frac45\alpha e^{-\alpha t} + \frac15\beta e^{-\beta t}\right)\right)^K.
$$
We then compute the probability of $\{\mathcal A\leqslant B\}$ by integrating over the joint density:
\begin{align}
\mathbb P(\mathcal A\leqslant B\} &= \int_0^\infty \int_0^\infty f_{\mathcal A}(s)f_{\mathcal B}(t)\mathsf 1_{s\leqslant t}\ \mathsf ds\ \mathsf dt\\
&=\int_0^\infty \int_0^t f_{\mathcal A}(s)f_{\mathcal B}(t)\mathsf 1_{s\leqslant t}\ \mathsf ds\ \mathsf dt\\
&= \int_0^\infty \int_0^t K \left(\frac{1}{5} \alpha ^2 e^{-\alpha s}\frac{4}{5} \beta ^2 e^{-\beta s}\right) \left(\frac{1}{5} \alpha  e^{-\alpha s}+\frac{4}{5} \beta  e^{-\beta s}\right)^{K-1}\cdot\\
&\quad\quad\quad\quad\quad\! K \left(\frac{4}{5} \alpha ^2 e^{-\alpha t}+\frac{1}{5} \beta ^2 e^{-\beta t}\right) \left(1-\frac{1}{5} 4 \alpha  e^{-\beta t}-\frac{1}{5} \beta  e^{-\beta t}\right)^{K-1}\mathsf ds\ \mathsf dt.
\end{align}
I was not able to evaluate this integral with Mathematica, however.
