I want to make sure my answer here is correct, I am extremely confident that it is but I am taking no chances :)
Question: The probability that a random bus arrives on time is $1 - (0.5)^5$. What is the probability that at least one of two buses arriving independently will arrive on time?
Solution:
P(At least one arrives on time) = 1 - P(Neither arrive on time)
= 1 - P(Bus 1 not on time $\cap$ Bus 2 not on time)
(By independence...)
= 1 - P(Bus 1 not on time) $*$ P(Bus 2 not on time)
Now the probability that a random bus is not on time
= 1 - P(Bus is on time)
= $1 - (1 - (0.5)^5)$
= $1 - (1 - \frac{1}{32}$)
= $\frac{1}{32}$
So
1 - P(Bus 1 not on time) $*$ P(Bus 2 not on time)
= 1 - $\frac{1}{32}\frac{1}{32}$
= $\frac{1023}{1024}$
Correct, yes?