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12 1. Nations of Probability
1.4.1 Calculus of Probability
Suppose that A and B are two arbitrary events. Here we summarize some of
the standard rules involving probabilities.
Additive Rule:
Conditional Probability Rule:
Multiplicative Rule:
The additive rule was proved in the Theorem 1.3.1, part (iv). The conditional
probability rule is a restatement of (1.4.1). The multiplicative rule follows
easily from (1.4.1). Sometimes the experimental setup itself may directly indi-
cate the values of the various conditional probabilities. In such situations, one
can obtain the probabilities of joint events such as A ∩ B by cross multiplying
the two sides in (1.4.1). At this point, let us look at some examples.
Example 1.4.3 Suppose that a company publishes two magazines, M1 and
M2. Based on their record of subscriptions in a suburb they find that sixty
percent of the households subscribe only for M1, forty five percent subscribe
only for M2, while only twenty percent subscribe for both M1 and M2. If a
household is picked at random from this suburb, the magazines publishers
would like to address the following questions: What is the probability that the
randomly selected household is a subscriber for (i) at least one of the maga-
zines M1, M2, (ii) none of those magazines M1, M2, (iii) magazine M2 given
that the same household subscribes for M1? Let us now define the two events
A: The randomly selected household subscribes for the magazine M1
B: The randomly selected household subscribes for the magazine M2
We have been told that P(A) = .60, P(B) = .45 and P(A ∩ B) = .20. Then,
P(A ∪ B) = P(A) + P(B) P(A ∩ B) = .85, that is there is 85% chance that
the randomly selected household is a subscriber for at least one of the
magazines M1, M2. This answers question (i). In order to answer ques-
c
c
tion (ii), we need to evaluate P(A ∩ B ) which can simply be written as 1
P(A ∪ B) = .15, that is there is 15% chance that the randomly selected
household subscribes for none of the magazines M1, M2. Next, to answer