 
Summary: It's dark and you're trying to open the door...
We begin with a very useful formula.
The multiplication rule. (See page 68 in the book.) Suppose E1, . . . , Em are events. Then
P(Em · · · E1) = P(EmEm1 · · · E1)P(Em1Em2 · · · E1) · · · P(E2E1)P(E1).
Proof. Just apply the definition of conditional probability and watch everything cancel.
Problem. Suppose it's dark and you have n keys to n different doors in your pocket and your trying to
open one of these doors. You take keys out of your pocket and try them until, as will surely happen, the
key opens the door. You're taking Mathematics 135 so you're pretty smart and you don't put a key that
doesn't work back in the pocket that originally had all the keys in it. Suppose k {1, . . . , n}. What is the
probability that the kth key pulled out works?
Solution. For each i {1, . . . , n} let Ei be the event that the ith key pulled out works and let Fi be the
complementary event. Applying the multiplication rule we find that
P(Ek) = P(Ek Fk1 · · · F1)
= P(EkFk1 · · · F1)P(Fk1Fk2 · · · F1) · · · P(F2F1)P(F1)
=
1
n  (k  1)
n  (k  2)  1
n  (k  2)
· · ·
