Let $X$ and $Y$ be two independent random variables (weibull distributed) and $Z=X+Y$. I am trying to find $\mathbb P\big(Z\geq z ~\cap~ X\leq x\big)$. I know that $$ \mathbb P(Z\geq z \cap X\leq x) = \mathbb P\big(Z\geq z\,\big|\,X\leq x\big)\cdot\mathbb P\big(X\leq x\big) $$ however I cannot calculate that conditional probability either. I can calculate $\mathbb P\big(X\leq x\big)$ (straightforward) and I can find an approximation of $\mathbb P\big(Z\leq z\big)$.
Anyobody has an idea on how to proceed? Thank you
Joan
$P\{X+Y \geq z, X \leq t\}$ (be aware of the slight change in notation) is the probability that the random point $(X,Y)$ lies above the line $x+y = z$ and to the left of the line $x=t$. So you can find the probability by integrating $f_{X,Y}(x,y)$ over this wedge-shaped region. This is a double integral in which it might be a tad easier to have the inner integral be with respect to $y$ and the outer with respect to $x$ but that is a matter of personal preference: integrating in either order should give you the same answer.
Addendum in response to OP's comment: If the integrals described above are difficult to evaluate but you have the exact value of $P\{X \leq x\}$ and a good approximation to $P\{X+Y \leq z\}$, then note that for $z \leq x$, the event $\{X+Y \leq z\}$ is a subset of the event $\{X \leq x\}$ and so $$P\{X+Y \geq z, X \leq x\} = P\{X \leq x\} - P\{X+Y \leq z\}, ~~ z \leq x.$$ For $z > x$, the right side of the above equation is a lower bound on the desired probability.