Probability of waiting a certain time for a phone booth

Question

I'm working on the following problem for my probability class.

Suppose you arrive at a public phone booth and find someone else using it. Suppose T is a random variable that represents the waiting time till you are able to get in. Suppose further that the mean length of a phone conversation is about 10 minutes.

a) Assume the following and find the probability that you have to wait at least 10 minutes before you get in. $$f_T(t)=\begin{cases} \frac{1}{10}e^{-t/10} \, \, \, \, t\ge 0 \\ 0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, otherwise \end{cases} $$

b) Find the probability that you have to wait between 10 and 20 minutes.

c) Suppose that the person talking has been in the booth for over 30 minutes. Find the probability that you still have to wait between 10 and 20 minutes.

Parts (a) and (b) were pretty easy, I'm just confused on how to set up part (c). I have a feeling I'd need to use conditional probability here, so I set it up like this originally:

$$ P(T>30 \, | \, 10 \le T \le 20)=\frac{P(T>30 \, \, \cap \, \, 10 \le T \le 20)}{P(10 \le T \le 20)} $$

But the numerator would, unless I'm mistaken, be the null set. $T>30$ and $10 \le T \le 20$ don't have any elements in common. Because of this, I believe I've made a mistake setting up this problem.

Any help would be appreciated, thank you.

Answer 1

You have the conditioning the opposite way around.

$P(A|B)$ means the probability that $A$ happens $given$ that $B$ happens. We're given that the waiting time is at least $30$ minutes and we wish to find the probability that we have to wait between $10$ and $20$ more minutes.

The exponential distribution is memoryless.

That is,

$$P(s+10<T<s+20 \mid T \gt s)=P(10\lt T\lt20)$$

where $s\geq30$

As a check of your work for $(a)$ and $(b)$, R statistical software gives the following:

$P(T\gt 10)$

> 1-pexp(10,1/10)
[1] 0.3678794

$P(10<T<20)$

> pexp(20,1/10)-pexp(10,1/10)
[1] 0.2325442