Ggtkuk

Guide:

1) Draw rectangle $2le xle 4,$ $3le yle 5$.

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 $7/2cdot 1/4=7/8$.

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 $B$:
$$E(3.7,3.3).$$

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) $$P(X) = P(X cap C) + P(X capoverline{C})=P(X | C) P(C) + P(X mid overline{C})P(overline{C})$$

Can you go on from here ?

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 $P(Z<0)$

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

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.

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.

The joint distribution for arrival times of A and B is
$$P(A,B)d!Ad!B = frac{1}{4}d!Ad!Bqquad 2<A<4, mathrm{and} ,3<B<5$$

We need the probability that A is less than B
$$P(A<B) = int_3^5 d!B int_2^{min(B,4)}d!A = frac{7}{8}$$