# Symmetry of Uniform Distribution PDF

I'm studying probability theory and came across an exercise problem that I couldn't quite understand, even with the solution, and was hoping someone could give me some insight.

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. on that day. What is the probability that Carl arrives first?

Apparently the answer is: "By symmetry, the probability that Carl arrives first is $$\frac{1}{3}$$." But I don't really know how this rather too simple solution came to be.

My initial approach is to divide cases as

$$P(C \lt A \lt B) + P(C \lt B \lt A)$$ and find each probability. However, wouldn't they differ depending on what time Carl arrives, and thus we would have to further condition?

Any feedback is appreciated. Thank you.

## 2 Answers

Let C be the event that Carl arrives first,B be the event that bob does and A the event that Alice does. Clearly, $$P(A)+P(B)+P(C)=1$$ since someone has to arrive first. Now notice that since they all pick their time arrival in the same way and independently, they should all have the same probability of arriving first. Thus, $$3P(C)=1$$

• Hello. Thanks for the swift reply. So, just to confirm that I understand correctly, because the three people have equally likely probability to arrive first, we could write it as $P(A) = P(B) = P(C)$, which would give us $P(A) = P(B) = P(C) = \frac{1}{3}$. Am I correct? – Seankala Nov 25 '18 at 3:12
• That’s exactly correct. They each have an equal chance of arriving first and the sum of their chances of arriving first must add up to 1 – Alex Nov 25 '18 at 4:45

Without loss of generality, consider a unit cube $$ABCDA_1B_1C_1D_1$$. Cut the cube along $$BB_1D_1D$$ and $$BCD_1A_1$$ to get the quadrilateral pyramid $$BAA_1D_1D$$, which has the volume: $$V=\frac13 \cdot S_{AA_1D_1D} \cdot AB=\frac13 \cdot 1^2 \cdot 1=\frac13.$$