A die is thrown twice and the number on each throw is recorded. There are
clearly 6 possible outcomes for the first throw and 6 for the second throw. By
the multiplication principle, there are 36 possible outcomes for the two throws.
If D is the event a double-six, then since there is only one possible outcome of
the two throws which leads to a double-six, we must have P(D) = 1=36.
Now let E be the event six on the first throw and F be the event six on the
second throw. We know that P(E) = P(F) = 1=6. If we are interested in the
event G, at least one six, then G = E [F, and using the addition law we have