Boole's inequality and uniform distribution Let $p_i, i = 1, 2, \cdots, n$ be random variables where each $p_i$ has a uniform distribution on $[0,1]$. Let $\alpha \in (0,1)$ be a constant. 

Show that $P(\min_{1 \le i \le n} p_i \le \alpha/n) \le \alpha$ using
  Boole's inequality.

I know that Boole's inequality states that $P(\cup_{n=1}^{\infty} A_n) \le \sum_{n=1}^{\infty} P(A_n)$ for any events $A_n$. However, how do I apply that inequality here?

A closely related problem is as follows. Assume now that $p_1, \cdots,
 p_n$ are jointly independent. Compute $P(\min_{1 \le i \le n} p_i \le
 \alpha/n)$ and its limit as $n$ approaches infinity.

Here, I know that the joint distribution is just the product of the individual uniform distributions, but I am unsure of how to calculate the probability and the limit required.
 A: 
Right, I was just thinking along those lines as well. Is my following train of thought correct? 

Yes.
The minimum of a set is no greater than a constant, exactly when at least one element of the set is not greater than the constant. Then we apply Boole's inequality.  Finally we call upon the fact that the random variables are uniformly distributed over $[0;1]$.
$$\begin{align}\mathsf P\left(\left(\min\limits_{1\leq i\leq n} p_i\right)\leq\dfrac \alpha n\right) ~&=~ \mathsf P\left(\bigcup\limits_{1\leq i\leq n}\left(p_i\leq\dfrac \alpha n\right)\right) \\[1ex] &\leq~ \sum\limits_{1\leq i\leq n}\mathsf P\left(p_i\leq\dfrac \alpha n\right)\\[1ex] &=~ n\cdot\frac \alpha n\\[2ex]\hline\therefore~\mathsf P\left(\left(\min\limits_{1\leq i\leq n} p_i\right)\leq\dfrac \alpha n\right) ~ &\leq~ \alpha\end{align}$$

On the other hand, we can use the complement.   The minimum of the set is greater than a constant, exactly when all of them are.   So, if the random variables are independent, then we use the definition of that. 
$$\begin{align}\mathsf P\left(\left(\min\limits_{1\leq i\leq n} p_i\right)\leq\dfrac \alpha n\right) ~&=~ 1- \mathsf P\left(\left(\min\limits_{1\leq i\leq n} p_i\right)\gt\dfrac \alpha n\right) \\[1ex] &=~ 1-\mathsf P\left(\bigcap\limits_{1\leq i\leq n} \left(p_i\gt\dfrac \alpha n\right)\right)\\[1ex] &=~\phantom{1-\prod\limits_{1\leq i\leq n}\mathsf P\left(p_i\gt\dfrac \alpha n\right)}\\[1ex] &=~\phantom{1-\prod\limits_{1\leq i\leq n}\left(1-\mathsf P\left(p_i\leq \dfrac \alpha n\right)\right)} \\[2ex]\hline\therefore~\mathsf P\left(\left(\min\limits_{1\leq i\leq n} p_i\right)\leq\dfrac \alpha n\right) ~&=~ \phantom{1-\left(1-\frac \alpha n\right)^n}\\[2ex]\therefore~\lim\limits_{n\to\infty}\mathsf P\left(\left(\min\limits_{1\leq i\leq n} p_i\right)\leq\dfrac \alpha n\right) ~&=~ \phantom{0}\end{align}$$
Note: This is not an inequality.   It is an equality.
