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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?

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

Let $B_ell$ be the event that $B$ arrives after $4$pm.

Let $C$ be the union : $C=A_e cup B_ell$.

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

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

Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$

Can you go on from here ?

Guide:

1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.

2) The total area of the rectangle and square is $5$, so pdf is $1/5$.

3) Find total area of the square and rectangle above the line, which is $4.5$.

5) Finally, the required probability is $4.5cdot 1/5=9/10$.

Here is the graph:

$hspace2cm$

Bus $A$ arriving before bus $B$:
$$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
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 $87.5$%.