1  Properties of conditional probability
Let E and F be events of a sample space S of an experiment, then we have
Property 1 P(S|F) = P(F|F) = 1
We know that
P(S|F) = P(S
F)
P(F)
1
P(F)
P(F)
∩
=
=
Also
P(F|F) = P(F
F)
P(F) 1
P(F)
P(F)
∩
=
=
Thus
P(S|F) = P(F|F) = 1
Property 2 If A and B are any two events of a sample space S and F is an event
of S such that P(F) ≠ 0, then
P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)
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MATHEMATICS
In particular, if A and B are disjoint events, then
P((A∪B)|F) = P(A|F) + P(B|F)
We have
P((A∪B)|F) = P[(A
B)
F]
P(F)
∪
∩
= P[(A
F)
(B
F)]
∩P(F)
∪
∩
(by distributive law of union of sets over  intersection)
=
P(A
F)+P(B
F)–P(A
B 
F)
P(F)
∩
∩
∩
∩
= P(A
F)
P(B
F)
P[(A
B) 
F]
P(F)
P(F)
P(F)
∩
∩
∩
∩
+
−
= P(A|F) + P(B|F) – P((A ∩B)|F)
When A and B are disjoint events, then
P((A ∩ B)|F) = 0
⇒
P((A ∪ B)|F) = P(A|F) + P(B|F)
Property 3 P(E′|F) = 1 − P(E|F)
From Property 1, we know that P(S|F) = 1
⇒
P(E ∪ E′|F) = 1
   since  S = E ∪ E′
⇒
P(E|F) + P (E′|F) = 1
    since E and E′ are disjoint events
Thus,
P(E′|F) = 1 − P(E|F)
Let us now take up some examples Example 1 If P(A) = 7
13 , P(B) = 9
13  and P(A ∩ B) = 4
13 , evaluate P(A|B) Solution We have 
4
P(A
B)
4
13
P(A|B)=
9
P(B)
9
13
∩
=
=
Example 2 A family has two children What is the probability that both the children are
boys given that at least one of them is a boy