492 CHAPTER^8 Discrete Probability
detail situations like repeated coin flipping, and we explain how to view events in separate
experiments as events in a single experiment.8.3.1 A Multiplication Principle
Suppose that a printer is out of order (down) 8% of the time and a photocopier is out of
order 10% of the time. Checking whether these machines are up or down can be regarded
as an experiment with sample space Q2 = {(0, 0), (1, 0), (0, 1), (1, 1)}, where the first and
second positions describe the state of the printer and the copier, respectively, and 0 and 1
denote down and up, respectively. Surely, (1, 1) and (0, 0) are not equally likely outcomes.
Hence, we seek another method for assigning a probability density.
We might reason about this situation as follows: The status of the photocopier and the
status of the printer have nothing to do with one another (assuming power failures or high
office temperatures did not put them both out-of-order at the same time). Therefore, the
photocopier could be down 10% of the time that the printer happens to be down and also
10% of the time that the printer happens to be up. Since the printer is down 8% of the time,
and 10% of the 8% that the copier is also down, it seems to be intuitively clear that 0.8% of
the time, both machines are down. The following example further illustrates this intuitive
reasoning. Later, we will prove this method assigns numbers that satisfy the definition of a
probability density function.Example 1. Suppose that you share a telephone with the other members of your house-
hold and that 30% of the time during the day, one of the others is using the telephone.
Suppose further that you wish to reach a certain service number. You know that 40% of
the time an incoming call to this service number is answered immediately, 35% of the time
the call is placed on hold in a queue, and 25% of the time you must call again, because the
queue is full and the number is busy. Based on intuitive reasoning, what is your estimate
of the probability that you can reach the service number from your home telephone with
no delay?Solution. Imagine two experiments X 1 and X 2 .Experiment X 1 consists of checking the
status of the telephone. We choose a sample space Q 1 with two outcomes, 0 and 1 for busy
and free, respectively, and we assign a probability density of Pl (0) = 0.3 and pl (1) = 0.7.
(The p is subscripted with a 1 to indicate that outcomes of Q21 are being considered.)
Experiment X 2 consists of checking the status of the service number. Here, we choose
Q 2 = {0, 1, 2} where 0 and I denote busy and answered immediately, respectively, and 2
denotes hold. We assign P2(0) = 0.25, p2(0) = 0.40, and p2(^2 ) = 0.35. Define the sam-
ple space Q2 = Q I X02:Q = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}Here, we have enumerated all possible pairs of outcomes, one from Q21 and one from Q22.
The outcome from Q21 is written as the first element of the pair, and the outcome from
Q2 2 is written as the second element. This product is just the product of sets defined in
Section 1.3.4.
If the home telephone is free 70% of the time and the status of the service number
is not related to the status of the home telephone, then we estimate that 40% of the time
that the home phone is free, the service number is answered immediately. Hence, p(1, 1)