# Probability that Carl has to wait at least 10 minutes for one of the others and for both of the others to show up?

This question is for the other subquestions for the same problem here. For those not willing to click the link, I will post the exercise problem here as well.

Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arrive independently at uniformly distributed times between 1 p.m. and 1:30 p.m.

For the rest of the problem, assume Carl arrives first at 1:10 p.m. and condition on this fact.

(b) What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?

(c) What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?

Apparently, my approach to the solution has given me the answers in a swapped manner, which I can't understand why. Allow me to elaborate.

My approach

Let's say that the events for Alice and Bob arriving are each $$A$$ and $$B$$ i.i.d. on Unif($$10,\ 30$$).

(b) The probability that Carl will wait more than 10 minutes for at least one of the others to arrive is:

(Probability that Carl waits less than 10 minutes for Alice and more than 10 for Bob) + (Probability that Carl waits less than 10 minutes for Bob and more than 10 for Alice) + (Probability that Carl waits more than 10 minutes for both)

which gives me the result of $$\left(\frac{1}{2} * \frac{1}{2}\right) + \left(\frac{1}{2} * \frac{1}{2}\right) + \left(\frac{1}{2} * \frac{1}{2}\right) = \frac{3}{4}$$.

(c) Probability that Carl waits more than 10 minutes for both is the probability that $$A$$ and $$B$$ both fall in the interval ($$20$$, $$30$$) in the total interval ($$10$$, $$30$$). Therefore we get the probability $$\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$$.

However, apparently the correct answer is $$\frac{1}{4}$$ for (b) and $$\frac{3}{4}$$ for (c), and I'm not entirely sure why.

The rationale the textbook gives is that for (b) we need to compute the time that both of them arrive after 1:20 p.m., and for (c) we simple take the complement that both of them arrive between 1:10 and 1:20 p.m. $$1 - \frac{1}{4}$$. I'm not entirely sure how these solutions make sense, though, and was hoping someone would be kind enough to provide me with some insight.

Thank you.

• The explanations in the book seem reversed to me. I get the same answers you do. Nov 25, 2018 at 4:04
• That's strange... I should bring this issue up with my instructor again, as the first time it was brushed off. May I ask, was the rationale that led you to your answer the same as mine?
– Sean
Nov 25, 2018 at 4:26
• Thanks. Then your solution looks good. Nov 25, 2018 at 4:36
• I did it exactly as the book describes, but for the other problems. I think the answers are simply reversed. Nov 25, 2018 at 6:57
• Thanks everyone. This is a bit ridiculous. I spent hours trying to figure out why the answer was like this, and also another good hour debating whether I'm fit for graduate studies or not and letting the existential doubt kick in. You guys saved me.
– Sean
Nov 25, 2018 at 7:01

It seems to me that questions (b) and (c) are both genuinely ambiguous, and both your solution and the one given in the textbook could be considered correct.

(b) What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?

I read this as asking for the probability that $$\ \min(A, B)>10\$$, where $$\ A\$$ is the time Carl has to wait for Alice to arrive and $$\ B\$$ the time he has to wait for Alice to arrive. While this seems to me to be the most natural way of reading the question, I recognise that it can also be read as asking for the probability that at least one of $$\ A\$$ and $$\ B\$$ is greater than $$10$$—that is, the probability that $$\ \max(A,B)>10\$$—and this may well seem to be the more natural reading to others.

(c) What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?

I initially read this as asking for the probability that Carl had to wait for more than $$10$$ minutes for both Alice and Bob to be present—that is, the probability that $$\ \max(A,B)>10\$$. However, it seems to me just as natural to take it as asking for the probability that both Alice and Bob take more than $$10$$ minutes to arrive—that is, the probability that $$\ \min(A,B)>10\$$.

So this is my understanding of the problem, the logic was annoying for me to understand as well.

b) So we can also define the event "Carl has to wait more than 10 minutes for one of them to arrive" as none of them arrived between the interval of 1:10-1:20 i.e none of them arrived within 10 minutes.

So we know that The probability of Alice and Bob have not arrived is 0.5 each. So we get the probability of 1/4 that none of them arrived within 10 minutes. From what I saw in your method you have also considered that one of them arrives within 10 minutes which shouldn't be considered in the problem since it's not at least! Otherwise your method was correct

c)similarly here we know that the probability of each arriving is 0.5 so both of them arriving within the next 10 minutes is 1/4.

If Carl needs to wait more than 10 minutes for them definitely both of them arrive between 1:20-1:30. so if the probability of them arriving before 1:20 is 1/4 then the event that they arrive between 1:20-1:30 corresponds with the event that they did not arrive at between 1:10-1:20 so if the probability that they arrive between 1:10-1:20 is 1/4 then probability that they did not is 3/4.

At least that is what I understood from their explanation I still think it is a bit odd but I this is what I interpretted.