I'm learning probability, specifically continuous random variables and distributions, and need help with the following problem :
The duration of a conversation on the phone follows the exponential distribution with an expected value of two minutes, i.e. $E[X] = 2'$. Find
$(1)$ The probability a conversation to last more than $4'$.
$(2)$ The probability that for $5$ conversations all of them last more than $4'$.
$(3)$ The probability that for $50$ conversations all of them last more than $4'$.
Since I'm having difficulties for $(2)$ and $(3)$, I'm going to share my work for $(1)$.
$(1)$ The CDF of the exponential distribution is given by $F(x) = P(X \leq x) = 1 - e^{-\lambda x}$ for $x > 0$, where the rate parameter $\lambda = 1/\mu = 0.5$. By the compliment rule we get that $$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\lambda x}) = e^{-\lambda x}.$$
Therefore, the requested probability is $$P(X > 4) = e^{(-0.5)(4)} = 0.1353.$$
Is my work correct for $(1)$? Any help for $(2)$ and $(3)$ will be much appreciated. Since we are asked to find the probability for more than one event I don't know how to apply a similar argument.
Correct for $1$.
For $2$, assume that $X_1, X_2, \ldots X_5$ are the durations of $5$ independent conversations. Then the required probability is $P(X_1>4, X_2>4, \ldots, X_5>4) = \prod_{i=1}^{5}P(X_i>4) = \prod_{i=1}^{5}e^{-4\lambda}$. The first equality follows from the independence assumption and second equality follows from the identical distributions of the variables.
Similarly proceed for $3$.