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Unit 3: Conditional Probability and Independence Baye’s Theorem

Let A , A and A , respectively denote the events that the first, second and third mangoes are Notes
1 2 3
good. Then P(A ) = 90/100, P(A | A ) = 89/99, and P(A | A  AZ) = 88/98 according to the
1 2 1 3 1
classichl definition. Thus.
90 89 88
P(A A A ) = . .  0.727.
1 2 3 100 99 98
We end this section with a derivation of a well-known theorem in probability theory, called the
Bayes’ theorem.
C
Consider an event B and its complementary event B . The pair (B, B ) is called a partition of ,
C
C
C
since they satisfy B  B = , and B B is the whole sample space . Observe.that for any
event A,
C
C
A = A = A(BB ) = (AB)(AB ).
Since AB and AB are subsets of the disjoint sets B and B , respectively, they themselves are
C
C
disjoint. As a consequence, P(A) = P(AB) + P(AB ).
C
Now using Relation (10), we have
C
Here we do not insist that P(B) and P(B ) be positive and follow the convention stated in
Remark 3.
It is now possible to extend Equation (11) to the case when we have a partition of  consisting of
more than two sets. More specifically, we say that the n sets B , B . . . . ,B constitute a partition
1 2 n
of  if any two of them are disjoint, i.e., 1
B B = , i  j, i, j = 1, . . . , n
i j
and their union is , i.e.,

n
 B   .
j
j 1

We can now write for any event A,
 n  n
A = A = A  B j   (A  B ).

  j
 j 1  j 1


Since AB and AB are respectively subsets of B and B, i  j, they are disjoint. Consequently
i j i j
by P7,
n
P(A) =  P(A  B )
j
j 1

n
or P(A) =  P(B )P(A  B ). ... (12)
j
j

j 1
which is obtained by using (10). This result (12) leads to the celebrated Bayes’ theorem, which we
now state.

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