Let $X$ and $Y$ be independent and identically distributed random variables with exponential distribution of parameter 2. Let $Z: = \min \{X, Y\}$ then what is $P (Z \leqslant4 \mid X> 3)$?
My attempt: $Z$ distributes exponential with parameter $4$, that is, $1-\exp \{4z\}$, then $F_z(4)=1-\exp\{16\}$. $X$ distributes exponential with parameter $2$, that is, $1-\exp \{2x\}$, then $P(x>3)=1-(1-\exp \{6\})=\exp \{6\}$. How can I use these facts?
Your approach is hard to use because $Z$ is independent of neither $X$ nor $Y$, so I'll do something different. Fixing $X=x>3$, the conditional probability becomes $1$ if $x\in(3,\,4)$, or $P(Y\le4)=1-e^{-8}$ if $x>4$. So$$P(Z\le4|X>3)=\frac{e^{-6}-e^{-8}+e^{-8}(1-e^{-8})}{e^{-6}}=1-e^{-10}\approx0.9999546.$$Python seems to agree, although it's hard to be sure at that precision level. (To make the code run faster, which is crucial when measuring probabilities just under $1$, I sampled the distribution to get $X-3$, exploiting the memorylessness of the Exponential distribution.)