p(blue) 5 24/100 5 .24, p(green) 5 16/100 5 .16, and so on. But what is the probability
that I will draw a blue or green M&M instead of an M&M of some other color? Here we
need the additive law of probability.
Given a set of mutually exclusive events, the probability of the occurrence of one event
or another is equal to the sum of their separate probabilities.
Thus, p(blue or green) 5 p(blue) 1 p(green) 5 .24 1 .16 5 .40. Notice that we have im-
posed the restriction that the events must be mutually exclusive, meaning that the occurrence
of one event precludes the occurrence of the other. If an M&M is blue, it can’t be green. This
requirement is important. About one-half of the population of this country are female, and
about one-half of the population have traditionally feminine names. But the probability that
a person chosen at random will be female orwill have a feminine name is obviously not.
50 1 .50 5 1.00. Here the two events are notmutually exclusive. However, the probability
that a girl born in Vermont in 1987 was named Ashley or Sarah, the two most common girls’
names in that year, equals p(Ashley) 1 p(Sarah) 5 .010 1 .009 5 .019. Here the names are
mutually exclusive because you can’t have both Ashley andSarah as your first name (unless
your parents got carried away and combined the two with a hyphen).The Multiplicative Rule
Let’s continue with the M&M’s where p(blue) 5 .24, p(green) 5 .16, and p(other) 5 .60.
Suppose I draw two M&M’s, replacing the first before drawing the second. What is the
probability that I will draw a blue M&M on the first trial anda blue one on the second?
Here we need to invoke the multiplicative law of probability.
The probability of the joint occurrence of two or more independent events is the prod-
uct of their individual probabilities.
Thus p(blue, blue) 5 p(blue) 3 p(blue) 5 .24 3 .24 5 .0576. Similarly, the probability of
a blue M&M followed by a green one is p(blue, green) 5 p(blue) 3 p(green) 5
.24 3 .16 5 .0384. Notice that we have restricted ourselves to independent events, mean-
ing the occurrence of one event can have no effect on the occurrence or nonoccurrence of
the other. Because gender and name are not independent, it would be wrong to state that
p(female with feminine name) 5 .50 3 .50 5 .25. However it most likely would be correct
to state that p(female, born in January) 5 .50 3 1/12 5 .50 3 .083 5 .042, because I know
of no data to suggest that gender is dependent on birth month. (If month and gender were
related, my calculation would be wrong.)
In Chapter 6 we will use the multiplicative law to answer questions about the independ-
ence of two variables. An example from that chapter will help illustrate a specific use of this
law. In a study to be discussed in Chapter Six, Geller, Witmer, and Orebaugh (1976) wanted
to test the hypothesis that what someone did with a supermarket flier depended on whether
the flier contained a request not to litter. Geller et al. distributed fliers with and without this
message and at the end of the day searched the store to find where the fliers had been left.
Testing their hypothesis involves, in part, calculating the probability that a flier would con-
tain a message about littering andwould be found in a trash can. We need to calculate what
this probability would be if the two events (contains message about littering and flier in
trash) are independent, as would be the case if the message had no effect. Ifwe assume that
these two events are independent, the multiplicative law tells us that p(message, trash) 5
p(message) 3 p(trash). In their study 49% of the fliers contained a message, so the proba-
bility that a flier chosen at random would contain the message is .49. Similarly, 6.8% of the
fliers were later found in the trash, giving p(trash) 5 .068. Therefore, if the two events are
independent, p(message, trash) 5 .49 3 .068 5 .033. (In fact, 4.5% of the fliers withSection 5.2 Basic Terminology and Rules 115additive law of
probability
multiplicative law
of probability