Suppose $X_1,X_2,X_3$ are independently and identically distributed with probability density function $f(x)$. Let event $A$ be "at least one random variable in $X_1,X_2,X_3$ is greater than $t$". What is $f(x|A)$?
I know that if the event is "$X_1>t$", then I can define the set $\{x:x>t\}$, and write $f(x|X_1>t)=\frac{f(x)\mathbf{1}(x\in \{x:x>t\} )}{Pr(X>t)}$, where $\mathbf{1}(\cdot)$ is the indicator function. I don't know how to extend this formula to find $f(x|A)$, because I don't see how to construct the set in the indicator function.
In general:$$f(x)=f(x|A)P(A)+f(x|A^c)P(A^c)\tag1$$and event $A^c=\{X_1,X_2,X_3\leq t\}$ is more convenient to work with than event $A$.
Note that:$$P\left(X_{1}\leq x,A^{c}\right)=\begin{cases} F\left(x\right)F\left(t\right)^{2} & \text{if }x\leq t\\ F(t)^3 & \text{otherwise} \end{cases}$$ so that:$$P\left(X_{1}\leq x\mid A^{c}\right)=\begin{cases} F\left(x\right)/F\left(t\right) & \text{if }x\leq t\\ 1 & \text{otherwise} \end{cases}$$
and:$$f\left(x|A^{c}\right)=\begin{cases} f\left(x\right)/F\left(t\right) & \text{if }x\leq t\\ 0 & \text{otherwise} \end{cases}$$ Applying $(1)$ we then find:$$f\left(x\mid A\right)=\frac{f\left(x\right)-f\left(x\mid A^{c}\right)P\left(A^{c}\right)}{P\left(A\right)}=\begin{cases} \frac{f\left(x\right)\left(1-F\left(t\right)^{2}\right)}{1-F\left(t\right)^{3}} & \text{if }x\leq t\\ \frac{f\left(x\right)}{1-F\left(t\right)^{3}} & \text{otherwise} \end{cases}$$