Probability of a $\max$ number of an array of Uniform rv's being higher than $x$ Recently I faced this question

Let $U_1, U_2, U_3$ all come from a uniform$(0,1)$ distribution. Let $M = \max(U_1, U_2, U_3)$. Estimate (to $3$ significant digits) the probability of $M > 0.75$.

So as a started, I started taking baby steps.
At first, I thought if I only had one value, and for that the probability would be $0.25\%$. Then I thought abut the addition rule and I figured this couldn't be right, as if I had $4$ elements the probability would be $100\%$, which is not correct. By reading the law of total probability, I couldn't think of how I could apply it with this question as well.
How should I approach this question in order to fully understand and solve it?
 A: Hint: If the $U_i$ are supposed to be independent, you can calculate
$$\begin{align*}P(M > 0.75) = &1 - P(M \le 0.75) = 1 - P(U_1 \le 0.75, U_2 \le 0.75, U_3 \le 0.75)\\
 =&1 - P(U_1 \le 0.75) P(U_2 \le 0.75)P(U_3 \le 0.75)\end{align*}$$
A: $P(M>=0.75)=1-P(M<0.75)$
$P(M<0.75)=P(U_1<0.75,U_2<0.75,U_3<0.75)$, indeed , if the maximum is smaller than 0.75, so are all of them , and reciprocally . 
Assuming they are mutually independent,
$ P(U_1<0.75,U_2<0.75,U_3<0.75)=P(U_1<0.75)P(U_2<0.75)P(U_3<0.75)$
$=P(U_1<0.75)^3$
$=0.75^3$
A: I assume that the $U_i$'s are independent. 


*

*Step 1: Make sure that you understand the following equivalence (if and only if statement) $$\max{\{U_1,U_2,U_3\}}\le 0.75\iff U_1\le 0.75,\;U_2\le 0.75,\;U_3\le 0.75$$ (the LHS implies the RHS and vice versa, can you see this?). This implies that 
$$\Pr\left(\max{\{U_1,U_2,U_3\}}\le 0.75\right)=\Pr\left(U_1\le 0.75,\;U_2\le 0.75,\;U_3\le 0.75\right)$$ 

*Step 2: The rv's $U_1, U_2, U_3$ have the same distribution which you can denote with $U$ (ok, this is pretty much a trivial step), so you can write the previous relation as $$\Pr(M\le 0.75)=\Pr(U\le 0.75)^3=0.75^3=0.422$$ But you want the probability of the complementary event $M>0.75$ hence $$0.422=\Pr(M\le 0.75)=1-\Pr(M>0.75)\implies \Pr(M>0.75)=0.578$$

