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72 CHAPTER 3 SOME BASIC PROBABILITY CONCEPTS
EXAMPLE 3.4.4
We wish to compute the joint probability of Early age at onset 1E2 and a negative family
history of mood disorders 1A2 from knowledge of an appropriate marginal probability and
an appropriate conditional probability.
Solution: The probability we seek is P(E ¨ A). We have already computed a marginal
probability, P(E) = 141>318 = .4434, and a conditional probability,
P1A ƒ E2 = 28>141 = .1986. It so happens that these are appropriate marginal
and conditional probabilities for computing the desired joint probability.
We may now compute P(E ¨ A) = P(E)P(A ƒ E) =1.443421.19862
= .0881. This, we note, is, as expected, the same result we obtained earlier
for P(E ¨ A).
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We may state the multiplication rule in general terms as follows: For any two events A
and B,
P1A ¨ B2 = P1B2P1A ƒ B2, if P1B2 Z 0
(3.4.1)
For the same two events A and B, the multiplication rule may also be written as
P1A ¨ B2 = P1A2P1B ƒ A2, if P1A2 Z 0.
We see that through algebraic manipulation the multiplication rule as stated in
Equation 3.4.1 may be used to find any one of the three probabilities in its statement if
the other two are known. We may, for example, find the conditional probability P1A ƒ B2
by dividing P1A ¨ B2 by P1B2. This relationship allows us to formally define conditional
probability as follows.
DEFINITION
The conditional probability of A given B is equal to the probability
of A º B divided by the probability of B, provided the probability of
B is not zero.
That is,
(3.4.2)
We illustrate the use of the multiplication rule to compute a conditional probability with
the following example.
EXAMPLE 3.4.5
P1A ƒ B2 =
We wish to use Equation 3.4.2 and the data in Table 3.4.1 to find the conditional probability,
P1A ƒ E2.
Solution: According to Equation 3.4.2,
P1A ¨ B2
, P1B2 Z 0
P1B2
P1A | E2 = P1A ¨ E2>P1E2
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