Two buses A and B arrive independently at a bus station at random rate of 5/hour and 4/hour respectively. A passenger comes to the bus station at 10 am. What is the probability that it takes at most 5 minutes before the first bus arrives at the station?
Let X and Y be random variables indicating time of arrivals of Bus A and B respectively. Then, X and Y are exponentially distributed random variables with rate 5 and 4 respectively. So the probability density functions of X and Y are 5*exp(-5*t) and 4*exp(-4*t).
Now, let Z = min(X,Y) be a new random variable. Then, I need to calculate P(Z<5/60).
Question: My solution sheet goes as follows: P(Z<5/60) = 1-P(Z>5/60) = 1-P(X>5/60, Y>5/60) and goes on like this. But why cannot I just calculate P(Z<5/60) = P(X<5/60, Y<5/60) directly? These two give me different answers, meaning my attempt might be wrong. Why do we need to calculate the complementary probability?