Let $X_{1},X_{2} $ be independent and $X_{1},X_{2} \sim U(1,6)$, $Z= \min(X_{1},X_{2})$, $ T = \sqrt{Z}$. Find $EZ$ and $P(T \le \frac{3}{2})$.
What I tried is the following : I took a geometric approach since I don't see a Continuous Random Variables approach with functions. So if out first point lands on the point $a$ then the second point should be in the interrval $6-a$, so our function is $Z = a(6-a)$ and then to find $EZ$ I will find the integral $$\int_{1}^{6} a^2(6-a) \,da$$
I am not sure if this is this is the way to solve it.
Here's a hint about the minimum of two random variables to get you started.
For any real-valued random variables $X_1$ and $X_2$, let $F_1(t)$ and $F_2(t)$ be their cumulative distribution functions, so that $1-F_1(t)$ is the probability that $X_1>t$ and similarly for $1-F_2(t)$. If $X_1$ and $X_2$ are independent, then the probability that both $X_1$ and $X_2$ are greater than $t$) is the product of these probabilities $(1-F_1(t))(1-F_2(t))$, and hence the probability that at least one of $X_1$ and $X_2$ is at most $t$ equals $1-(1-F_1(t))(1-F_2(t))$. This is therefore the cumulative distribution function of $\min\{X_1,X_2\}$.
In our case, we have $F_1(t)=F_2(t)=\tfrac15(t-1)$ (restricted to the interval $[1,6]$ of course). Therefore the cumulative distribution function of $Z = \min\{X_1,X_2\}$ is $$ 1-\big(1-\tfrac15(t-1)\big)\big(1-\tfrac15(t-1)\big) = \tfrac1{25}(-t^2 + 12 t -11), $$ and its density function is therefore the derivative, which is $\tfrac2{25}(-t+6)$.