Exponential random variable with Minimum and Maximum Probability If $X_1, X_2, X_3, X_4, X_5$ are independent and identically distributed exponential random variables with the parameter λ, compute
(a) $P{(min(X_1,...,X_5) \le a});$
(b) $P{(max(X_1,...,X_5) \le a}).$
I am real lost on what the question is even asking here. I know how to work with exponential random variables in terms of $f(x)=λe^{-λx}$ and $F(x)=1-e^{-λx}$. But how would I use those ideas here?
 A: Hints: 
For part (a), you can express the probability as
$$P{(min(X_1,...,X_5) \le a}) = 1-P((X_1>a) \cap (X_2>a) \cap (X_3>a) \cap (X_4>a) \cap (X_5>a))$$
For part (b), you have
$$P{(max(X_1,...,X_5) \le a}) = P((X_1 \leq a) \cap (X_2 \leq a) \cap (X_3 \leq a) \cap (X_4 \leq a) \cap (X_5 \leq a))$$
Based on the fact that the random variables are i.i.d. exponential distributed, can you take it from here?
A: Let $X$ be exponentially distributed with parameter $\lambda$ with distribution function F.
Using the fact that.
$$
\max(X_{1},\dotsc, X_{5}) \leq a \iff X_{1} \leq a, X_{2} \leq a,\dotsc, X_{5} \leq a 
$$
$$
P(\max(X_{1},\dotsc, X_{5})\leq a ) = \prod_{i=1}^5P(X_{i}\leq a) = F(a)^5
$$
where in the first equality we use independence. In the second, the fact that they are identically distributed.
For the second one use the fact that
$$
\min(X_{1}, \dotsc, X_{5}) \ge a  \iff X_{1} \geq a, X_{2} \geq a,\dotsc, X_{5} \geq a 
$$
to compute $P(\min(X_{1}, \dotsc, X{5}) \ge a )$  using similar reasoning as above.
