Let $~X,Y∼U_{[0,1]}~$ independently. Find $P(\max(X,Y)≥0.8∣\ \min(X,Y)=0.5)$ Let  $X,Y∼U_{[0,1]}$  independently. Find  $$P(\max(X,Y)≥0.8∣\ \min(X,Y)=0.5)~.$$ 
 A: I plotted here, zone where both the inequalities you mention in your question are verified. It should give you a good starting point for an answer.


*

*The blue lines comes from: 
$\min(X,Y) = 0.5 \Leftrightarrow (X = 0.5 \text{ and }  Y > 0.5) \text{ or } (Y = 0.5 \text{ and }  X > 0.5)$

*the second lines come from $\max(X,Y) \ge 0.8 \Leftrightarrow X \ge 0.8 \text{ or }  Y \ge 0.8$

A: Let $Z=\min\{X,Y\}$ and $W=\max\{X,Y\}$.
For $0\leq z\leq w\leq1$, we have $$P(W\leq w)=w^2,$$ and $$P(Z>z,W\leq w)=P(X>z,Y>z,X\leq w,Y\leq w)=(w-z)^2,$$ so  $$P(Z\leq z,W\leq w)=P(W\leq w)-P(Z>z,W\leq w)=w^2-(w-z)^2.$$ Differentiate to obtain the density function $f_{Z,W}(z,w)=2$, $0\leq z\leq w\leq1$.
Use a similar method to obtain the density function $f_{Z}(z)$ of $Z$, which is $f_{Z}(z)=2(1-z)$. So $$f_{W|Z}(w|z)=\frac{1}{1-z},\ \ \ 0\leq z\leq w\leq1.$$ Therefore 
$$P(W>0.8|Z=0.5)=\int_{0.8}^1f_{W|Z}(w|0.5)\,dw=\int_{0.8}^12\,dw=\frac{2}{5}.$$
A: The solution of the diligent bonehead:
Let $U=\min(X,Y)$ and $V=\max(X,Y)$.The probability in question can be calculated the following way
$$P(V≥0.8∣\ U=0.5)=\int_{0.8}^1 f_{V\mid U=0.5}(v)\ dv=\int_{0.8}^1\frac{f_{U,V}(0.5,v)}{f_U(0.5)}dv.$$

First we calculate the common cdf of $U$ and  $V$:
$$F_{U,V}(u,v)=P(U<u\cap V<v)=$$
$$P(\min(X,Y)<u\cap \max(X,Y)<v)=$$
$$=P([X<u\cup Y<u]\cap [X<v\cap Y<v])$$
assuming that $0\leq u,v \leq 1$.
Now
$$ [X<u\cup Y<u]\cap [X<v\cap Y<v])=$$
$$=[X<u\cap X<v\cap Y<v]\cup[Y<u\cap X<v\cap Y<v].$$
That is,
$$P([X<u\cup Y<u]\cap [X<v\cap Y<v])=$$
$$P([X<u\cap X<v\cap Y<v]\cup[Y<u\cap X<v\cap Y<v])=$$
$$=P(X<u\cap X<v\cap Y<v)+P(Y<u\cap X<v\cap Y<v)-P(X<u\cap X<v\cap Y<u\cap Y<v)=$$
$$=P(X<\min(u,v))P(Y<v)+P(Y<\min(u,v))P(X<v)-P(X<\min(u,v))P(Y<\min(u,v))=$$
$$=\begin{cases}
2uv-u^2&\text{ if }&u\leq v\\
0&\text{ if }&u> v
\end{cases}$$
assuming that $u,v\in [0,1]$.
This is the common cdf. 
The common pdf is $2$ within the lower half triangle of the unit square and $0$ otherwise. We can get that by taking the partial dericative of the cdf with respect to $u$ and then with respect to $v$ or vice versa.
That is,
$$f_{\min(X,Y),\max(X,Y)}(u,v)=2$$
if $u,v$ are positive, less that $1$, and $u\leq v$.
Second, let's calculate the cdf of $U=\min(X,Y)$:
$$P(\min(X,Y)<u)=P(X<u\cup Y<u)=$$
$$=P(X<u)+P(Y<u)-P(X<u\cap Y<u)=$$
$$=2u-u^2$$
between $0$ and $1$,  zero below $0$ and $1$ above $1$. The pdf is the derivative of the same. That is,  
$$f_{\min(X,Y)}(u)=2-2u$$
Then
$$f_{\min(X,Y),\max(X,Y)\mid \min(X,Y)}(u,v)=\frac{f_{\min(X,Y),\max(X,Y)}(u,v)}{f_{\min(X,Y)}(u)}=$$
$$=\frac1{1-u}$$
if $(u,v)\in$ the triangle described above. At $u=0.5$ if $0\leq v\leq 1$ :
$$ f_{\min(X,Y),\max(X,Y)\mid \min(X,Y)}(u,v)=2.$$
Finally
$$P(\max(X,Y)≥0.8∣\ \min(X,Y)=0.5)=\int_{0.8}^1f_{\min(X,Y),\max(X,Y)\mid \min(X,Y)=0.5}(0.5,v)dv=$$
$$=\int_{0.8}^1 2 dv=2×0.2=0.4.$$
A: Applying the Law of Total Probability over the partition of which variable is the minimum, and then applying the independence of the random variables, gives us the evaluation:
$$\begin{align}\mathsf P(\max\{X,Y\}\geq 0.8\mid\min\{X,Y\}=0.5)&={{\mathsf P(X\geq 0.8, X\geq 0.5\mid Y=0.5)}+{\mathsf P(Y\geq 0.8, Y>0.5\mid X=0.5)}}\\[1ex]&={{\mathsf P(X\geq 0.8\mid Y=0.5)}+{\mathsf P(Y\geq 0.8\mid X=0.5)}}\\[1ex]&=\mathsf P(X\geq 0.8)+\mathsf P(Y\geq 0.8)\\[1ex]&=0.2+0.2\\[1ex]&=0.4\end{align}$$
