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?

I assume that the $U_i$'s are independent.

1. 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 $$\Pr\left(\max{\\{U_1,U_2,U_3\\}}\le 0.75\right)=\Pr\left(U_1\le 0.75,\;U_2\le 0.75,\;U_3\le 0.75\right)$$
2. 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 $$0.422=\Pr(M\le 0.75)=1-\Pr(M>0.75)\implies \Pr(M>0.75)=0.578$$