# Probability of waiting time

Question: At a railroad junction, a car and a truck arrive between 7:15 and 7:30. A train stops the traffic for five minutes from 7:20.

What is the probability that the car and truck waited for three minutes or more?

My understanding of this problem is that the total area is (15)(5) = 75. I have the Y axis from 0 to 15 (representing 7:15 to 7:30) and the X axis from 0 to 5 (representing 7:20 to 7:25)

I believe there is a small triangle that represents the area of interest, with vertices at (3,5), (3,7), and (5,5). So this area is (.5)(2)(2) = 2.

Therefore I have a final probability of $(\frac{2}{75})^2$.

Any help would be much appreciated.

• The question is not clear. Do you need P(car and truck both wait for 3 minutes or more) ? Nov 19, 2015 at 9:30
• Either vehicle will have to wait 3 or more minutes if it arrives between 7:20 and 7:22 Nov 19, 2015 at 9:33
• Yeah my mistake, P(car and truck both wait for 3 minutes or more). Nov 19, 2015 at 9:39
• For one vehicle probability is $\frac{2}{15}$. So for both it should be $(\frac{2}{15})^2$ (assuming independence) Nov 19, 2015 at 9:42

Assume independent uniform distribution of the car and truck arrival, and using time indexing from minute $0$ to $15$ (instead of $7:15 - 7:30$). For $A$ the event of arrival (of either car or truck) we have that $A$ ~ $Uniform(0,15)$. Because it will be enough if car and truck arrive between $5$ and $7$ (between $7:18$ and $7:22$), It follows that
(1) $\mathbb{P}(wait(A)\geq3) = \frac{7-5}{15} = \frac{2}{15}$
(2) $\mathbb{P}(max\{wait(A_1), wait(A_2)\} \geq 3) = \mathbb{P}(wait(A)\geq3)^2 = (\frac{2}{15})^2$
Thus, the sought after answer (if I interpreted the exercise correctly) is $(\frac{2}{15})^2$, like already pointed out in the comments.