# Determine the expected value and the probability of some events (applicative probability problem)

Suppose that the time that a phone call lasts is a random variable with density function given by:

$$f(t) = \begin{cases} \frac{1}{5} e^{-\frac{1}{5}t} & \text{if } t > 0 \\ 0 & \text{another case} \end{cases}$$ I want to first determine some probabilities for the call:

1. Takes longer than 5 minutes. We have, by the apparent continuity of the random variable

$$P(X >5)= \int_{5}^{\infty} \frac{1}{5} e^{-\frac{1}{5}t}=[-e^{-\frac{1}{5}t}]_{5}^{\infty}=e^{-1}$$

2. Takes between 5 and 6 minutes.

$$P(5 \leq X \leq 6)= \int_{5}^{6} \frac{1}{5} e^{-\frac{1}{5}t}=[-e^{-\frac{1}{5}t}]_{5}^{6}=e^{-1}-e^{-\frac{6}{5}}$$

3. Takes less than 5 minutes.

$$P(X< 5)= \int_{0}^{5} \frac{1}{5} e^{-\frac{1}{5}t}=[-e^{-\frac{1}{5}t}]_{0}^{5}=1-e^{-1}$$

4. Takes less than 6 minutes given that it took at least 3 minutes

Here note that if $$A=\{X<6 \}$$ and $$B=\{X\geq 3 \}$$, then $$A \cap B = \{3 \leq X <6 \}$$, as $$P(B) = \int_{3}^{\infty} \frac{1}{5} e^{-\frac{1}{5}t}=e^{-\frac{3}{5}}$$ and $$P(A \cap B)=\int_{3}^{6} \frac{1}{5} e^{-\frac{1}{5}t}=-e^{-\frac{6}{5}}+e^{-\frac{3}{5}}$$

Then

$$P\{X<6|X \geq 3 \}=\frac{-e^{-\frac{6}{5}}+e^{-\frac{3}{5}}}{e^{-\frac{3}{5}}}=1-e^{-\frac{3}{5}}$$

But now, if $$C(t)$$ be the amount that must be paid by the user for every call that lasts $$t$$ minutes

$$C(t)=\begin{cases} 500 & \text{if } 0

And I want to find the mean cost of a call, then I can consider

\begin{align*} E(C(t)) &= \int_{-\infty}^{\infty} |C(t)|f(t) & = & \int_{0}^{5} 500 \frac{1}{5} e^{-\frac{1}{5}t} & + & \int_{5}^{10} 750 \frac{1}{5} e^{-\frac{1}{5}t} & + & \int_{10}^{\infty} 100t \frac{1}{5} e^{-\frac{1}{5}t} \\ & & = & 100(5-5e^{-1}) & + & e^{-2}(750e-750) & + & 1500e^{-2} \end{align*}

What I have indicated is correct? was there any correction? any help is welcome!!

• There's at least one typographical error in (3).
– whuber
Nov 11, 2023 at 19:59
• @whuber You're right, thank you, it was less than 5 minutes. Nov 11, 2023 at 20:18