The probability of Bus A arriving before Bus B

Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?

My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?

• Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct? – Vimath Mar 23 at 4:10
• Yes, the arrival of buses is independent events. – IrinaS Mar 23 at 4:45
• As a nitpick, the words "probability" and "odds" are not interchangeable. They are related, yes, but they do not mean the same thing. The probability of picking an ace from a well shuffled standard deck is $\frac{1}{13}$. The odds however are $1:12$ for, or equivalently $12:1$ against. If you only ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(A\mid B')$, not $P(A/B')$ – JMoravitz Mar 23 at 13:59
• @JMoravitz I missed the syntax there , I'll try not to repeat it. – Vimath Mar 23 at 14:53

Guide:

1) Draw rectangle $$2\le x\le 4,$$ $$3\le y\le 5$$.

2) The area of the rectangle is $$4$$, so pdf is $$1/4$$.

3) Draw line $$y=x$$.

4) Find area of the rectangle above the line, which is $$7/2$$.

5) Finally, the required probability is $$7/2\cdot 1/4=7/8$$.

Here is the graph:

$$\hspace{2cm}$$

Bus $$A$$ arriving before bus $$B$$: $$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$ Bus $$A$$ arriving after bus $$B$$: $$E(3.7,3.3).$$

• do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify? – IrinaS Mar 23 at 5:15
• The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square. – Craig Hicks Mar 23 at 8:18
• Thanks, I rolled back to my first answer. – farruhota Mar 23 at 8:29
• But the graph was great! Bring back the graph! – Craig Hicks Mar 23 at 8:32
• my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will. – farruhota Mar 23 at 8:35

First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.

You’re mistaken when you say that if you know Bus A is arriving between $$3$$ and $$4$$, then there is an equal probability of either bus arriving first. There is a $$50$$% probability that Bus B arrives after $$4$$. The probabilities are equal only if you know that Bus A arrives between $$3$$ and $$4$$ and also that Bus B arrives between $$3$$ and $$4$$. That parlay occurs only $$25$$% of the time.

So $$75$$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $$25$$% of the time. Thus, the probability that Bus A arrives first is $$87.5$$%.

Let $$A_e$$ ($$A$$ early) be the event that bus $$A$$ arrives before $$3$$pm.

Let $$B_\ell$$ ($$B$$ late) be the event that bus $$B$$ arrives after $$4$$pm.

Let $$C$$ be the union : $$C=A_e \cup B_\ell$$. Hence $$P(C)=P(A_e)+P(B_\ell)-P(A_e \cap B_\ell)=P(A_e)+P(B_\ell)-P(A_e)P(B_\ell)$$

Let $$X$$ be the event of interest ( bus $$A$$ arrives before bus $$B$$).

What we know (don't we?) is that $$P(X | C)=1$$ and $$P(X | \overline{C})=0.5$$

Then we can write (total probability) $$P(X) = P(X \cap C) + P(X \cap\overline{C})=P(X | C) P(C) + P(X \mid \overline{C})P(\overline{C})$$

Can you go on from here ?

• Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4. – IrinaS Mar 23 at 5:26
• @IrinaS - "A has already arrived before $B_\ell$ happens" -- that would be $P(B_\ell | A_\ell)$. And you're right -- if $A_\ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_\ell$ and $B_\ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A \cup B)$ = 1 – cag51 Mar 23 at 7:23
• @IrinaS What cag51 says above. – leonbloy Mar 23 at 14:01

Define a new variable $$Z = A-B = A+ (-B)$$.

Whenever $$Z<0 \Rightarrow A (A arrives before B)

Since A and (-B) are independent, the pdf of Z is the convolution of the pdfs of A and (-B): $$f_Z(z) = f_A(a)*f_{-B}(b)$$.

Solving the convolution graphically you get that: $$f_Z(z) = \cases{ \frac{z+3}{4}, -3 \leq z < -1 \\ \frac{1-z}{4}, -1 \leq z < 1}$$

Now compute $$P(Z<0)$$

• What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check. – Craig Hicks Mar 23 at 22:37

$$\int_{s=2}^4 \int_{t=3}^5 p(a=s) p(b=t) \delta(s

is the 2-d continuous integral equation. $$\delta(s is 1 when $$s and 0 otherwise. $$\delta(s bisects the total area into two parts, the sum of which is 1.

It is easily visualized and solved with a 2 x 2 hours block of time with a diagonal line $$\delta(s cutting through one corner.

The graph in farruhota's answer shows it.

• I wouldn't say "bisects". In other contexts, bisection implies equal parts. – David K Mar 23 at 14:40
• @DavidK - I agree. "partitions" is a much better word choice. – Craig Hicks Mar 23 at 22:06

What you missed in your approach to the question is that there is nothing in the problem statement that prevents bus $$B$$ from arriving after $$4$$ pm, and in fact (assuming uniform, independent distributions of the arrival times) half the time bus $$B$$ will arrive after $$4$$ pm, and in that case bus $$A$$ will have arrived first.

Likewise, there is nothing that requires bus $$A$$ to arrive after $$3$$ pm so that bus $$B$$ has a chance to arrive first. Bus $$A$$ can just as likely arrive before $$3$$ pm.

Let's take a frequentist approach. Suppose that these two buses run on this random schedule seven days a week, every day of the year. Let's watch them arrive for a few days and see what happens. Here's one possible way this might unfold:

• On Monday, bus $$A$$ arrived at $$2{:}38$$ and bus $$B$$ arrived at $$3{:}02$$. Although bus $$B$$ arrived almost as quickly as it possibly can, bus $$A$$ still arrived first.

• On Tuesday, bus $$A$$ arrived at $$3{:}05$$ and bus $$B$$ arrived at $$3{:}42$$. Bus $$A$$ arrived first.

• On Wednesday, bus $$A$$ arrived at $$2{:}50$$ and bus $$B$$ arrived at $$4{:}11$$. Bus $$A$$ arrived first.

• On Thursday, bus $$A$$ arrived at $$3{:}57$$ and bus $$B$$ arrived at $$4{:}30$$. Bus $$A$$ arrived first even though it was almost as late as it can be.

• On Friday, bus $$A$$ arrived at $$2{:}05$$ and bus $$B$$ arrived at $$4{:}56$$. Bus $$A$$ arrived first.

• On Saturday, bus $$A$$ arrived at $$3{:}19$$ and bus $$B$$ arrived at $$3{:}17$$. Finally we have observed an event in which bus $$B$$ arrived first!

In the long run, if we keep track of the relative frequency of days like Monday (when bus $$A$$ arrives before $$3$$ pm and bus $$B$$ arrives between $$3$$ and $$4$$ pm), we'll find that the relative frequency approaches $$1/4$$ of all the days. To put it simply, in the long run $$1/4$$ of the days will be like Monday.

Similarly, in the long run $$1/4$$ of the days will be like Wednesday and Friday, when bus $$A$$ arrived before $$3$$ pm and bus $$B$$ arrived after $$4$$ pm. Another $$1/4$$ of the days will be like Thursday, when bus $$A$$ arrived between $$3$$ and $$4$$ pm but bus $$B$$ arrives after $$4$$ pm.

That leaves just $$1/4$$ of the days in the long run when bus $$A$$ and bus $$B$$ both arrive between $$3$$ and $$4$$ pm. One half of those days ($$1/8$$ of all days in the long run) will be like Tuesday, when bus $$A$$ arrived first, and the other half of those days ($$1/8$$ of all days in the long run) will be like Saturday, when bus $$B$$ arrived first.

And that accounts for all possibilities. In one case, which happens $$1/8$$ of the time, bus $$B$$ arrives first. In all other cases bus $$A$$ arrives first.

• Agree with your logic. This is a very good and simple explanation. Thank you, @David K. – IrinaS Mar 23 at 18:30

The joint distribution for arrival times of A and B is $$P(A,B)d\!Ad\!B = \frac{1}{4}d\!Ad\!B\qquad 2

We need the probability that A is less than B $$P(A