Stats train probability I'm having trouble with my stats. The question says that the probability of the Amsterdam train arriving on time is .93, and the probability of the Brussels train arriving on time is .89.
The probability both trains are late is .05.
It asks what the probability that at least one train is on time. I'm not sure what to do. I drew circles representing the probabilities, but I'm not certain what the intersection value is supposed to be. Also, is it correct to say that P(A complement) U P(B complement) = .05?
I think that the complement of P(at least one train arrives on time) is P(no train arrives on time). So I think I need to subtract that probability from 1. 
I don't know how to get it, though.
 A: We see that there are four possibilities:
Case 1: Both trains are on time.
Case 2: Train 1 is on time, train 2 is late.
Case 3: Train 2 is on time, train 1 is late.
Case 4: Both trains are late.
You should see that the probabilities of each of the cases should add to $1$. (since these are the only possible outcomes). You are given that the probability that both trains are late is $0.05$, so the probability of one of the other three cases is $1-0.05=0.95$. We see that the other three cases can be summarized as "at least one train is on time", therefore the probability that at least one train is on time is $0.95$.
A: Denote with $A$ the event that the Amsterdam train arrives on time. Denote with $B$ the event, that the Brussels train arrives on time. You know that $P(A)=0.93$ and $P(B)=0.89$. With $A^c$ denoting the complement of the event $A$, you also know that
$P(A^c \cap B^c)=0.05$,
i.e. the probability of both trains arriving late (not on time) is 0.05.
You want to determine $P(A \cup B)$, i.e. the event that at least one of the trains arrive on time.
You can use that $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B)$, such that $P(A \cup B) = 1 - P(A^c \cap B^c) = 1 - 0.05 = 0.95$. Note that you only needed the probability of both trains arriving late! The tricks applied above and other rules can be found here: http://stattrek.com/probability/probability-rules.aspx.
For a picture of what is going on:

