The probability of Bus A arriving before Bus B

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

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 ?