Conditional Probability of Random Variable

Question

Question: Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ =1/8.If someone arrives immediately ahead of you at a public telephone booth, find the probability that

A) you will have to wait more than 5 minutes.

C) If you have waited for 5 minutes, what is the probability that you will have to wait more than 15 minutes in total?

My Solution (for A): P(X > 5) = e^-5/8 = 0.53

My Solution (for C): P(X > 10 | X >= 5) = P(X > 10 AND X >= 5) / P(X >= 5)

Using Bayes Rule,

= ( P(X >= 5 | X > 10) * P(> 10) ) / P(X >= 5)
= P(X > 10) / P(X >= 5)

Using exponential random variable rule: P(X > x) = e^(-λx)

= (e^-10/8) / (e^-5/8)
= 0.53

My Issue: I am not sure if I am using the correct equation for P(X >= 5). I don't understand how the probabilities can be the same after a condition has been placed.

I don't understand how you could have the same probability after a condition has been placed. Can you please help find an issue with my logic?

Answer 1

The expression $P(X > 10 \mid X \geq 5)$ does not do what you apparently think it does. The variable $X$ has to represent the same thing in both places where it occurs in this formula. If you keep using the definition from part A, then $X$ is the total amount of time you wait from the moment you arrive until you are able to make your call, in which case the condition in part C is that $X \geq 5$ and the event "you will have to wait more than $15$ minutes in total" is $X > 15.$ The event $X > 10$ would be "you have to wait more than $10$ minutes in total."

You seem to want $X > 10$ to be the probability that you wait an additional ten minutes. But that's redefining what $X$ is from the definition you used in part A. If you redefine $X$ in this way, then $X$ is really "total waiting time minus $5$ minutes," and the condition that you have waited at least $5$ minutes becomes $X \geq 0$.

So what you calculated is actually the probability that you have to wait an additional five minutes given that you already waited five minutes. And an interesting property of an exponential distribution for a waiting time is that it has no "memory": no matter how long you wait, the conditional probability of waiting another $N$ minutes is the same as the initial probability that you would wait $N$ minutes.

Answer 2

Your solution for A) is correct.

For C) you have tried to calculate the probability that you will have to wait more than $10$ minutes in total.

You could do so by stating that: $$P(X>10\mid X>5)=P(X>10\wedge X>5)/P(X>5)=P(X>10)/P(X>5)=e^{-\frac58}$$

No rule of Bayes is applied.

You were asked however to find the probability on waitinb $15$ minutes in total: $$P(X>15\mid X>5)=P(X>10)=e^{-\frac{10}8}$$

The exponential distribution is memoryless (which is nicely illustrated by the former calculation where it is actually shown that $P(X>10\mid X>5)=P(X>5)$). I make use of that fact.