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?
 A: 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 ?
A: 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 $B$:
$$E(3.7,3.3).$$
A: 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 $87.5$%.
A: Define a new variable
$Z = A-B  = A+ (-B)$. 
Whenever $Z<0 \Rightarrow A<B$ (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 $P(Z<0)$
A: 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.
A: $$\int_{s=2}^4 \int_{t=3}^5 p(a=s) p(b=t) \delta(s<t) $$ 
is the 2-d continuous integral equation.  $\delta(s<t)$ is 1 when $s<t$ and 0 otherwise.  $\delta(s<t)$ 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 $\delta(s<t)$ cutting through one corner. 
The graph in farruhota's answer shows it.  
A: The joint distribution for arrival times of A and B is
$$P(A,B)d\!Ad\!B = \frac{1}{4}d\!Ad\!B\qquad 2<A<4\, \mathrm{and} \,3<B<5$$
We need the probability that A is less than B
$$P(A<B) = \int_3^5 d\!B \int_2^{\min(B,4)}d\!A = \frac{7}{8}$$
