Every day, a comes to the bus stop at the same time, and boards the first bus that arrives. the arrival of the first bus is an exponential random variable with expectation 20 minutes. every day and independently, b arrives at the same bus stop at a random time uniformly distributed between a's arrival time and a's arrival time plus 30 minutes. what is the probability that a and b meet at the bus stop?

So the waiting time for a bus has density f (t) = λe-λtf (t) = λe-λt, where λλ is the rate. To understand the rate, you know that f (t) dtf (t) dt is a probability, so λλ has units of 1/[t]1/[t]. Thus if your bus arrives rr times per hour, the rate would be λ=rλ=r. Since the expectation of an exponential distribution is 1/λ1/λ, the higher your rate, the quicker you'll see a bus, which makes sense.

So define X=min (B1, B2) X=min (B1, B2), where B1 B1 is exponential with rate 33 and B2 B2 has rate 44. It's easy to show the minimum of two independent exponentials is another exponential with rate λ1 + λ2 λ1 + λ2. So you want:

P (X>20 minutes) = P (X>1/3) = 1-F (1/3), P (X>20 minutes) = P (X>1/3) = 1-F (1/3),

where F (t) = 1 - e - t (λ1 + λ2) F (t) = 1 - e - t (λ1 + λ2).